StarFury - bad physics or stupid Axis?
Axis
Apologies if this was discussed way back in the dark ages of season 1.
OK, so I was re-watching some S3 eps the other day (just confirming
that they were crap (-; )
Anyway, I got to Severed Dreams and I was watching the StarFury launch
sequence. And I got to thinking about the physics involved.
Now, as far as I can see, the Furies face away from the center of B5's
rotation, and when they are 'dropped' their motion is along a radius
extending from that center.
Time for some crap ASCII 'art' :
^
|L
|
|
_-ŻŻŻ-_
/ \
| * C |
\ /
Ż-___-Ż
----> R
If C is the center of rotation , R is the direction of rotaion of B5,
then L is the launch vector of the unpowered StarFury.
But, thinking about the analogy of swinging a ball on the end of a
string around your head and cutting the string, shouldn't the
StarFuries launch thusly?
L<------
_-ŻŻŻ-_
/ \
| * C |
\ /
Ż-___-Ż
----> R
Either I am missing something monumentally obvious here or there's a
grievous error in the physics of StarFury launches. Which is it?
If it turns out to be the latter, I never want to here anyone slagging
off S:AAB about the Hammerheads 'banking' in space again, OK? ;-)
//================= THIS WAS A MESSAGE FROM AXIS =================\\
// "What are you people doing standing around? /| \ / O /ŻŻŻ` \\
// I want you out there, bombing Chigs before / | \/ \__ \\
\\ they've had their morning coffee and donuts!" /__|__/\ | \ //
\\ COL. McQUEEN, 58th SQUADRON 'WILD CARDS' / | \_|____/ //
\\=================== <*> ====================/Ax...@netcomuk.co.uk//
Axis
je...@cam.ac.uk (James Broughton) felt sure that it was vital for our
survival to know that:
<snip>
>Ok, here goes. From Fall of Night, we know that the outside of the station is
>rotating a 60mph or about 27m/s. So when the Starfuries are lauched they would
>have a sidewides velocity of 27m/s. However this quickly becomes insignificant
>as even accelerating at only 4g (From Fall of Night also, we know that pilots
>take up to 10g before blacking out), the Starfury will have a forwards(radial)
>velocity of 39m/s after only 1 second and after 3 seconds moving fowards at
>about 120m/s. The sideways velocity would hardly be noticable even if, as I
>have assumed, the starfuries were not accelerating slightly off radial to
>counteract the sideways movement.
>All in all they would appear to come straight out although they would actually
>follow a curved path (parabolic path?).
Dang it! It's so hard to explain this with ASCII.
I accept your explanation would be correct apart this major problem:
If you watch some of the launch sequences in which the POV of the
observer is behind the SF for example Severed Dreams, you'll see that
there's a good few seconds after the SF is 'dropped' during which the
SF is unpowered, yet it somehow acquires a radial velocity away from
the station. *Then* the thrusters kick in. This can't be correct.
Also, given that the SF launch is unpowered, how do the SFs clear the
launch bay in the manner shown? They should be moving *sideways* and
then go 'BOOM!' as they are clipped by the walls of the launchbay.
In fact, the SD shot is a perfect one to analyse. There's a shot after
the SFs launch where they are flying, *unpowered*, directly towards
the camera, and you can see the launch bays in the background. They
remain directly behind the SFs in a perfect radial line.
More crappy ASCII art:
\
\ Launchbay remains 'behind' SF instead of rotating away
oooo
center o XX,, Vector of unpowered SF POV of observer
* of o XX> -----> <-- o>
rot. o XX``
oooo
/ | Launchbay should be moving in this direction
/ |
v
This is definitely wrong, and it's contradicted about thirty seconds
later when we see more SFs launch from a high external shot, and the
launch bays revolve away from behind the SFs. I'm afraid it's all
completely wrong. Check out the SD sequences, watch them as a
physicist, and I'm sure you'll agree.
Oh well, at least it's on-topic. ;-)
>>If it turns out to be the latter, I never want to here anyone slagging
>>off S:AAB about the Hammerheads 'banking' in space again, OK? ;-)
>OK. So S:AAB wins hands down on realism. :-)
>Maybe not.
The funny thing is, if you watch closely you can see the little
thruster clusters (great name!) fire on the Hammerheads, and they fire
in a manner consistent with performing banks, i.e. simulating a
centripetal force by varying the thrust linearly. The question is, why
would you *want* to do that instead of using StarFury-type
manoeuvreing?
Oh yeah, it looks good. Anyway the Centauri fighters (the
crescent-shaped ones) do it too - check out the S3 title sequence.
//================= THIS WAS A MESSAGE FROM AXIS =================\\
// "What are you people doing standing around? /| \ / O /秤畔 \\
Mr P.M. Martins Ferreira
Axis (Ax...@netcomuk.co.uk) wrote:
StarFury - bad physics or stupid Axis?
Axis - watch it, you're giving people ammunition...
: But, thinking about the analogy of swinging a ball on the end of a
: string around your head and cutting the string, shouldn't the
: StarFuries launch thusly?
: L<------
: _-ŻŻŻ-_
: / \
: | * C |
: \ /
: Ż-___-Ż
: ----> R
: Either I am missing something monumentally obvious here or there's a
: grievous error in the physics of StarFury launches. Which is it?
Well, I suggest you make the experiment. Are you trying to tell me that
if you cut the string the ball will be launched tangentially?!?!? NO!
It will have a tangential speed component, and a vertical one. So, the
end trajectory is the sum of both, the Starfuries distance themselves
from the station but (untill they change speed in the tangent direction)
continue aligned with the Cobra Bays. So: I think you're missing something
monumentally obvious. Hope this helped.
Aaaah, physics threads! Don't you just love'em?!?!?
Pedro
rupert smith
>Now, as far as I can see, the Furies face away from the center of B5's
>rotation, and when they are 'dropped' their motion is along a radius
>extending from that center.
>But, thinking about the analogy of swinging a ball on the end of a
>string around your head and cutting the string, shouldn't the
>StarFuries launch thusly?
<diagram snipped with fury going off tangentially>
>Either I am missing something monumentally obvious here or there's a
>grievous error in the physics of StarFury launches. Which is it?
aaaaaaaaah! <pauses to recover from sudden mania>
but as observed from the launching bay, as it rotates, the starfury is
moving radially away from the bay, just as an external observer sees the
fury move tangetially away from the launch ring. there is a problem with
rotation but i think that we can cope with that.
------------------------------------------------------------------------------
rupert smith linc...@sable.ox.ac.uk http://users.ox.ac.uk/~linc0015
And what rough beast, its hour come round at last,
Slouches towards Bethlehem to be born? -- "The Second Coming"
Axis
ferr...@liverpool.ac.uk (Mr P.M. Martins Ferreira) felt sure that it
was vital for our survival to know that:
>Axis (Ax...@netcomuk.co.uk) wrote:
>StarFury - bad physics or stupid Axis?
>Axis - watch it, you're giving people ammunition...
I'm used to being shot at.... especially by you! ;-)
If I'm proven wrong, I'll happily concede, but I don't think I *am*
wrong.
>: But, thinking about the analogy of swinging a ball on the end of a
>: string around your head and cutting the string, shouldn't the
>: StarFuries launch thusly?
>: L<------
>: _-ŻŻŻ-_
>: / \
>: | * C |
>: \ /
>: Ż-___-Ż
>: ----> R
>: Either I am missing something monumentally obvious here or there's a
>: grievous error in the physics of StarFury launches. Which is it?
>Well, I suggest you make the experiment. Are you trying to tell me that
>if you cut the string the ball will be launched tangentially?!?!? NO!
>It will have a tangential speed component, and a vertical one. So, the
>end trajectory is the sum of both, the Starfuries distance themselves
>from the station but (untill they change speed in the tangent direction)
>continue aligned with the Cobra Bays. So: I think you're missing something
>monumentally obvious. Hope this helped.
To be honest, I don't understand what you mean. Yet more crappy ASCII:
Side view:
*-------------\ Stone moving in circle parallel with the ground,
\O coming 'out of the page'
x|
x|
/ \
/ \
Plan view:
_-ŻŻŻ-_
| / \
| *---O |
V \ /
Ż-___-Ż
Right. Now, discounting gravity effects (which I can safely do because
there is no equivalent force acting along the rotation axis of the
station), S shows the direction that the stone would move if the
string was cut and F shows the direction that the StarFuries move:
\ * --------->S
\O
side x| In this view, the direction of the StarFuries, F, is
view x| perpendicular to the screen, coming straight out towards
/ \ you
/ \
_-ŻŻŻ-_ Think of this as a cross section
plan / \ of B5 perpendicular to the long
view | O | axis.
\ | /
Ż-_|_-Ż
*-------->S
|
|
|
V
S
Now, when you translate this to B5, you'll see that the motion is all
wrong. If the Furies were firing thrusters at launch, I could accept
this, but they don't. Just watch the Fury launch sequence before the
Sheridan 'hologram' speech in Severed Dreams to see what I mean.
>Aaaah, physics threads! Don't you just love'em?!?!?
Yup. At the end of the argument, you're either right or wrong (at
least in simple Newtonian physics). Much more satisfying than
aesthetic arguments, which usually end up with 'it sux' , 'no it
ROOLZ' , 'SUX!!' , 'ROOLZ BUTTMUCH!!!!!!!' - BOOM!
Interesting to see how this one pans out.
//================= THIS WAS A MESSAGE FROM AXIS =================\\
// "What are you people doing standing around? /| \ / O /ŻŻŻ` \\
Ian Pitcher
Ax...@netcomuk.co.uk (Axis) wrote:
<snip>
>The funny thing is, if you watch closely you can see the little
>thruster clusters (great name!) fire on the Hammerheads, and they fire
>in a manner consistent with performing banks, i.e. simulating a
>centripetal force by varying the thrust linearly. The question is, why
>would you *want* to do that instead of using StarFury-type
>manoeuvreing?
We sort of discussed this last night at the NE Gathering. My theory is
that because the craft are meant as "indo/exo atmospheric" fighters,
you would not want to have to fly them in two different ways (normal
lift & bank IN atmosphere / fight inertia OUT of atmosphere).
Therefore the planes use a computer controlled fly-by-wire system to
'simulate' atmospheric flight whilst in space (eg pull back on the
stick - nose goes up [due to thruster cluster])
Hope that makes the kind of sense I want it to make...
Ian
--
i.c.p...@ncl.ac.uk [+44 (0)191 2227322] [Orange +44 (0)973 638957]
http://claret.ncl.ac.uk/ http://www.ncl.ac.uk/~nicp/
A thought crossed his mind, looked for traffic, saw none & crossed back
Sion Arrowsmith
In article <5c5h1o$uk$1...@taliesin.netcom.net.uk>,
Axis <Ax...@netcomuk.co.uk> wrote:
>ferr...@liverpool.ac.uk (Mr P.M. Martins Ferreira) felt sure that it
>was vital for our survival to know that:
>>Axis (Ax...@netcomuk.co.uk) wrote:
>>StarFury - bad physics or stupid Axis?
>>: string around your head and cutting the string, shouldn't the
>>: StarFuries launch thusly?
>>: L<------
>>: / \
>>: | * C |
>>: \ /
>>: ~-___-~
>>: ----> R
>
>>: Either I am missing something monumentally obvious here or there's a
>>: grievous error in the physics of StarFury launches. Which is it?
>
>>Well, I suggest you make the experiment. Are you trying to tell me that
>>if you cut the string the ball will be launched tangentially?!?!? NO!
>>It will have a tangential speed component, and a vertical one. So, the
>>end trajectory is the sum of both, the Starfuries distance themselves
>>from the station but (untill they change speed in the tangent direction)
>>continue aligned with the Cobra Bays. So: I think you're missing something
>>monumentally obvious. Hope this helped.
>
>To be honest, I don't understand what you mean. Yet more crappy ASCII:
(Only this time it's mine. And, Axis, don't put high-bit-set
characters into your ASCII art.)
On these diagrams, the 'Fury is marked with a X and the bay by a
C (for cobra). This is the situation at launch:
L<-----X
_-~C~-_
/ \
| * |
\ /
~-___-~
----> R
This is the situation some time later, if the pilot hasn't fired
their thrusters:
L<X-----
_-~~~-_
C \
| * |
\ /
~-___-~
----> R
Now, as you can see, the 'Fury isn't really "above" the bay, but
it's pretty close. And anyway, this situation would never arise --
the 'Fury's thrusters would have been fired long ago, while it was
still nearly "above" the bay. The mathematically-inclined may wish
to mutter about approximating the sine of a small angle to the
angle as measured in radians at this point. Basically, you seem
to be forgetting that the station keeps on rotating (unlike the
string in the ball-and-string analogy).
As for how the 'Fury's get out of the bay in the first place, that's a
much better question. I think the docking clamps don't just drop them,
but have them slide along a short way. This would provide just enough
radial velocity to carry them clear.
--
\S -- si...@chiark.greenend.org.uk -- http://www.chiark.greenend.org.uk/~siona/
___ | Spot the sleeve notes: #5 (on a live album): `Some additional
\X/ | recording took place not a thousand miles away from the home of
<*> | the artiste. The generic term of this process is "cheating".'
Axis
ferr...@liverpool.ac.uk (Mr P.M. Martins Ferreira) felt sure that it
was vital for our survival to know that:
<explanation and pretty ASCII snipped for bandwidth>
I hate to bother you, but....
The moment the grapple is released, there are NO forces acting on the
Fury (except gravitational attraction to the station's mass - too
small to worry about here).
Call me a kipper, but I seem to recall that Newton's Laws say that, if
the net forces acting on a body are zero, then the velocity of that
body remains constant.
Therefore, at the moment the grapples are released, the Fury will
continue on whatever velocity vector it had at the moment the forces
became zero, i.e. at the moment of grapple release.
At that moment, the velocity of the Fury is tangential, i.e.
perpendicular to the radius of rotation. The unpowered,
zero-force-acting-on-it Fury will continue in that tangential velocity
until some other force acts upon it (i.e. thrusters or the launchbay
wall WHAP!) and *not* in a perfectly radial velocity as portrayed.
So I agree that, while the Fury is grappled to the station, it
experiences force F = m V^2/R as you say; unfortunately that's only
relevant when discussing the motion of a grappled Fury. As soon as
it's cut loose, it no longer applies.
>I actually think that a better example is Galileo's experiment. Let a
>ball fall from a Tower. If the Earth is rotating, why does the ball
>fall vertically? Simple: because, at the instant of lauch, the ball has
>the same rotation speed of Earth. Therefore, is "follows" Earth's
>rotation precisely untill it reaches the ground. Perfect vertical. But,
>for an outside observer, resting in relation to the Sun, for example,
>the ball's movement would have seemed a parabol.
Not convinced. It's the same principle as I outlined above. The only
reason the ball in the tower is rotating at the same frequency as the
Earth is that the tower is (to put it *very* crudely) a string. The
moment you drop it from the tower (cut the string) it ceases to
'rotate' about the Earth and is acted upon by Earth's gravity, plus
some negligible wind resistance.
To human perception, it might as well be falling straight, but it's
really a parabola (constant velocity tangential to the Earth at the
moment of release plus a perpendicular acceleration due to gravity).
Make the tower very tall, and the parabola will be very evident. Make
the tower tall enough, and the ball will never hit the Earth at all.
>THERE! Did that do it? If not, tell me, I'll try again. And again.
>And how many times it takes.
<unnecessary and unwarranted patronising received>
Keep at it.
</unnecessary and unwarranted patronising received>
Come on you physicists, you're being very quiet.....
//================= THIS WAS A MESSAGE FROM AXIS =================\\
// "What are you people doing standing around? /| \ / O /秤畔 \\
Zathras
I think to some extent that Axis is both right and wrong, there *is*
something dodgy about star fury launches, but it's not as bad as
Axis makes out.
ferr...@liverpool.ac.uk (Mr P.M. Martins Ferreira) scribbled
> Axis (Ax...@netcomuk.co.uk) wrote:
> : >Axis - watch it, you're giving people ammunition...
>
> : I'm used to being shot at.... especially by you! ;-)
>
> You're being unfair...
>
> : If I'm proven wrong, I'll happily concede, but I don't think I *am*
> : wrong.
I don't think you're wrong about the movement of either the station
or the Star Furies, but you are wrong with respect to the *apparent*
view of the scene. But first I want to dispel a myth that seems to be
cropping up here.
> Babylon 5 rotates across an axis, this rotation "creates" artificial
> gravity on the surface of the cilinder that is the station. Any object
Right!
> in that cilinder feels a force pushing it outwards, radially. With me
Wrong! See next bit!
<snip>
> Now, the centrifugal force the Fury is sensing will change its radial
This is where you're going wrong! Centrifugal force *does not exist* it
is an apparent *effect* of centripetal force. (Centripetal force being
the one that pulls you inward)
The station's skin can be considered equivalent to the person holding a
piece of string, and the fury as the ball attached to the string. The
person holding the string is providing a rotational force with an
angular velocity corresponding to 100 wibbles per sec at the Cobra bays.
This means that the *speed* (not velocity) of the ball is also 100
wibbles per sec. If the string is cut, the rotational force is *gone*.
Now a rotational force is one which has some component acting at 90
degrees to the angle of motion, and the force must point towards the
*centre* of rotation, hence the force that creates the rotation is a
centripetal force. However the person who is in the star fury will feel
a (subjective view) centrifugal force that seems to be pushing them away
from the station. To say that centrifugal force is a real *force* would
be the same as saying that the person who is pushed into the door of
their car when they accelerate too fast round a corner is exerting a
force on the car, when it is in fact the other way round.
Where am I going with this? Axis's diagram is correct, once the Star Fury
is dropped, it has no external forces acting upon it, and must therefore
most in a straight line from it's previous position.
X= Star Fury, R=Rotation direction, @= launch bay.
At Drop: Shortly after:
X X<------
_=*@*=_ _=***=_
/ \ /@ \
| | | |
\ / \ /
--___-- --___--
-----> R -------> R
BUT! If you're in a Star Fury, you're launch bay has rotated at the same
speed behind you, meaning that the launch bay should from a subjective
point of view should be line up with the centre of B5's rotation.
ie X POV --------> BAY | <*> core shuttle bit.
This is not exact since there *will* be an error, but even after a 60
degree rotation, the line up is still close. But there is a different
problem when you reach that angle, which is incorrect, which Axis seems
to have spotted. Assuming no external forces, then the Star Fury shouldn't
be rotating in the *same* direction as the station, which means that from
the last diagram, the orientation should (sorta ;) be:
ie + POV --------> BAY | <*> core shuttle bit.
As it were, since if you look at the before/after diagram, the X does
not rotate wrt to the overal cross-section, but should seem to an
observer on the station appear to rotate in a different direction from
the station.
Hence a third person view looking down the end of the station would see:
X<------
_=***=_
/@ \
| |
\ /
--___--
A person on the station in the star fury bay should seen the Star Fury
moving radially from the station, but lagging slightly behind the station,
until about 60 degrees has past, and then they should seem accelerate
in the opposite direction to the station's apparent motion.
ie they would think the launch path looked like:
X
/
/
/
/
/
/
/
|
|
_=*@*=_
/ \
| |
\ /
--___--
Also the Star Fury should seem to rotate towards the station - ever so
slowly. A large *speed* (say 1000 wibble's per sec) would tend to
exacerbate all these problems - probably causing the pilot to get dizzy,
and this would seem to account for the Cobra bay launchers being in the
bit joining the two spheroid bits together as close to the centre of the
station as possible. Not only would this allow for easier positioning
in the launch bays, but would reduces the launch speed which is a function
of angular velocity and distance from centre of rotation.
The fact that we don't see the slight (apparent) rotation of the Fury
wrt the station may be due to this small distance, and also Foundation
loved to move the camera - which would disguise any small rotation
or lack there-of.
I've always thought *cool* ever since I saw Star Furies launch, and still
think it's one of the most practical launch mechanisms we seen - except
for all the Shadow fighters launched in one go in The Long Twilight
Struggle.
My brain hurts. BAD Brain... <Tvvt..>
Z.
Who did AS-Level Physics cos he couldn't understand the math, but he
could play with his balls...
James Broughton
>OK, so I was re-watching some S3 eps the other day (just confirming
>that they were crap (-; )
>
>Anyway, I got to Severed Dreams and I was watching the StarFury launch
>sequence. And I got to thinking about the physics involved.
>
>Now, as far as I can see, the Furies face away from the center of B5's
>rotation, and when they are 'dropped' their motion is along a radius
>extending from that center.
>
>Time for some crap ASCII 'art' :
> ^
> |L
> |
> |
> _-ŻŻŻ-_
> / \
> | * C |
> \ /
> Ż-___-Ż
> ----> R
>
>If C is the center of rotation , R is the direction of rotaion of B5,
>then L is the launch vector of the unpowered StarFury.
>
>But, thinking about the analogy of swinging a ball on the end of a
>string around your head and cutting the string, shouldn't the
>StarFuries launch thusly?
> L<------
> _-ŻŻŻ-_
> / \
> | * C |
> \ /
> Ż-___-Ż
> ----> R
>
>Either I am missing something monumentally obvious here or there's a
>grievous error in the physics of StarFury launches. Which is it?
>
Ok, here goes. From Fall of Night, we know that the outside of the station is
rotating a 60mph or about 27m/s. So when the Starfuries are lauched they would
have a sidewides velocity of 27m/s. However this quickly becomes insignificant
as even accelerating at only 4g (From Fall of Night also, we know that pilots
take up to 10g before blacking out), the Starfury will have a forwards(radial)
velocity of 39m/s after only 1 second and after 3 seconds moving fowards at
about 120m/s. The sideways velocity would hardly be noticable even if, as I
have assumed, the starfuries were not accelerating slightly off radial to
counteract the sideways movement.
All in all they would appear to come straight out although they would actually
follow a curved path (parabolic path?).
>If it turns out to be the latter, I never want to here anyone slagging
>off S:AAB about the Hammerheads 'banking' in space again, OK? ;-)
OK. So S:AAB wins hands down on realism. :-)
Maybe not.
James
Ben Argyle
On 22 Jan 1997, James Broughton wrote:
> In article <5c548m$o1p$1...@taliesin.netcom.net.uk>,
> Ax...@netcomuk.co.uk (Axis) wrote:
>=20
> >OK, so I was re-watching some S3 eps the other day (just confirming
> >that they were crap (-; )
> >
> >Anyway, I got to Severed Dreams and I was watching the StarFury launch
> >sequence. And I got to thinking about the physics involved.
> >
> >Now, as far as I can see, the Furies face away from the center of B5's
> >rotation, and when they are 'dropped' their motion is along a radius
> >extending from that center.
> >
> >Time for some crap ASCII 'art' :
> > ^
> > |L
> > |
> > |
> > _-=AF=AF=AF-_
> > / \
> > | * C |
> > \ /
> > =AF-___-=AF
> > ----> R
> >
> >If C is the center of rotation , R is the direction of rotaion of B5,
> >then L is the launch vector of the unpowered StarFury.
> >
> >But, thinking about the analogy of swinging a ball on the end of a
> >string around your head and cutting the string, shouldn't the
> >StarFuries launch thusly?
> > L<------ =20
> > _-=AF=AF=AF-_
> > / \
> > | * C |
> > \ /
> > =AF-___-=AF
> > ----> R
> >
> >Either I am missing something monumentally obvious here or there's a
> >grievous error in the physics of StarFury launches. Which is it?
> >
>=20
>=20
> Ok, here goes. From Fall of Night, we know that the outside of the statio=
n is=20
> rotating a 60mph or about 27m/s. So when the Starfuries are lauched they =
would=20
> have a sidewides velocity of 27m/s. However this quickly becomes insignif=
icant=20
> as even accelerating at only 4g (From Fall of Night also, we know that pi=
lots=20
> take up to 10g before blacking out), the Starfury will have a forwards(ra=
dial)=20
> velocity of 39m/s after only 1 second and after 3 seconds moving fowards =
at=20
> about 120m/s. The sideways velocity would hardly be noticable even if, as=
I=20
> have assumed, the starfuries were not accelerating slightly off radial to=
=20
> counteract the sideways movement.
>=20
> All in all they would appear to come straight out although they would act=
ually=20
> follow a curved path (parabolic path?).=20
>=20
> >If it turns out to be the latter, I never want to here anyone slagging
> >off S:AAB about the Hammerheads 'banking' in space again, OK? ;-)
>=20
> OK. So S:AAB wins hands down on realism. :-)
> Maybe not.
>=20
> James
On the "I might be missing something obvious" slant again. Aren't the=20
'Furies launched from the top of the station on the bit that doesn't=20
move. I.e. the bit that makes up the forward cargo nacelles and stretches=
=20
back all the way down the station. Hence the relative rotational motion of=
=20
the station to the 'Furies is zero (I think/hope).
Go ahead, tell me I'm wrong. I've just come out of finals exam and my=20
mind's a bit wibbly (-:
--Ben
"Live Faust, die Jung"
__________________________________________________________________________=
___
|"It's nice to be important, | Email: B...@cs.man.ac.uk =
|
| but more important to be nice."| Homepage: http://www.cs.man.ac.uk/~argyl=
eb |
|________________________________|_________________________________________=
___|
| Abbot, Manchester Chapter of the Monks of Cool since MCMXCVI a.d. =
|
|__________________________________________________________________________=
___|
Mr P.M. Martins Ferreira
: I'm used to being shot at.... especially by you! ;-)
You're being unfair...
: If I'm proven wrong, I'll happily concede, but I don't think I *am*
: wrong.
I, on the other hand, am absolutely sure you ARE wrong. I'll try to
explain it a little better.
: To be honest, I don't understand what you mean. Yet more crappy ASCII:
My fault, I'll try to be more explicit. Here goes:
Babylon 5 rotates across an axis, this rotation "creates" artificial
gravity on the surface of the cilinder that is the station. Any object
in that cilinder feels a force pushing it outwards, radially. With me
so far? Now, the starfuries are stuck to the station, right? ASCII
follows:
\ You can take * to be the "graple" that's holding the
\ starfury, and S to be the starfury itself. Now, this
\ system is rotating at a speed V, along an axis perpendicular
\ to the sreen. This means that the starfury is "feeling"
| a centrifugal force, pushing him outside the station, of
|
*S F = m V^2/R
|
| with R the radius of the Cobra bays and m the fury's mass.
/ In the moment the "graple" releases it, this is the force
/ the fury is acted by, IN THE RADIAL DIRECTION. Which means,
/ the Starfury's velocity will only be changed in the radial
direction, its tangential velocity (V, it's the velocity
of rotation - not the angular one!) will remain unchanged.
Now, the centrifugal force the Fury is sensing will change its radial
speed, making him take distance from the station. The reason the Fury doesn't
crash into the bay doors (as you very correctly point out, there are shots
where we see the Furies being "dropped" and they stay aligned with the
Bay doors; they DO NOT activate their thrusters while inside the Bay) is that,
in the tangential direction, the Fury and the Bay door (the whole station
surface, actually) are moving in precisely the same direction, at precisely
the same speed. That is, tangentially, the fury is immobile with respect
to the Bay door. For sometime, of course. If the distance INSIDE THE
BAY the starfury had to go was quite big, it would crush against a wall.
But for that amount? Forget it.
In your string analogy: the stone's movement in the string's end has two
components :
- One, radial, due to the tension of the string, the distance between the
stone and the center of rotation is kept constant. The stone, on the
other hand, is acted by a centrifugal force, pushing it outwards - that's
why the string is stretched.
- Another, tangential, a component of velocity that is tangent to the
trajectory of the stone at each moment.
t=0 - the stone is released (for a perfect analogy with the Fury's launch,
you have to assume that the string is still rotating, at exactly the same
speed)
Scheme for t<0
pulled outwards by the centrifugal force
<--------<STONE>*--------> pulled this way by the string's
| tension
|
V Speed of rotation, or tangential
t=0: the tension disappears:
<--------<STONE>* ------------------The string keeps rotating
|
|
V Tangential
For some instant t>0, but VERY close to zero: the stone moved x m in the
outward direction, pushed by its acceleration, and y m in the tangent
direction. BUT: given that the string's tangent velocity is the same as that
of the stone, the string's extremity is stil "aligned" with the stone.
<--------<STONE>* -------------The string keeps rotating
|
|
V Tangential
I actually think that a better example is Galileo's experiment. Let a
ball fall from a Tower. If the Earth is rotating, why does the ball
fall vertically? Simple: because, at the instant of lauch, the ball has
the same rotation speed of Earth. Therefore, is "follows" Earth's
rotation precisely untill it reaches the ground. Perfect vertical. But,
for an outside observer, resting in relation to the Sun, for example,
the ball's movement would have seemed a parabol.
THERE! Did that do it? If not, tell me, I'll try again. And again.
And how many times it takes.
Pedro
Julian Barkway
=> Apologies if this was discussed way back in the dark ages of season 1.
=>
=> OK, so I was re-watching some S3 eps the other day (just confirming
=> that they were crap (-; )
=>
=> Anyway, I got to Severed Dreams and I was watching the StarFury launch
=> sequence. And I got to thinking about the physics involved.
=>
=> Now, as far as I can see, the Furies face away from the center of B5's
=> rotation, and when they are 'dropped' their motion is along a radius
=> extending from that center.
I've always had a problem with that. How can they be 'dropped' when B5
itself is the CoG? It's not that much better if it's assumed the SFs are
'dropped' from the zero-g part of B5...
...but why let a simple matter like the laws of physics get in the way of a
cool piece of CGI, eh?
============================================================================
Julian Barkway, |"Fred Davis, the doyen of snooker, now 67 years
jbar...@xeuronet.nl,| of age and too old to get his leg over, prefers
Amsterdam. | to use his left hand" -Ted Lowe, snooker commentator
============================================================================
Address deliberately corrupted (remove 'x's) to prevent unwanted email spam.
Jonathan Silverlight
In article <5c548m$o1p$1...@taliesin.netcom.net.uk>, Axis (Ax...@netcomuk.co.uk) writes:
>Apologies if this was discussed way back in the dark ages of season 1.
>
It's certainly been worked to death on rec.arts.sf.tv.babylon5
Axis
Zathras <zat...@mad.scientist.com> felt sure that it was vital for
our survival to know that:
>I think to some extent that Axis is both right and wrong, there *is*
>something dodgy about star fury launches, but it's not as bad as
>Axis makes out.
>I don't think you're wrong about the movement of either the station
>or the Star Furies, but you are wrong with respect to the *apparent*
>view of the scene. But first I want to dispel a myth that seems to be
>cropping up here.
Ah, you spotted it too! It seems Mr. Ferreira lives in a universe with
different Newtonian laws to the rest of us.
>ferr...@liverpool.ac.uk (Mr P.M. Martins Ferreira) scribbled
>> Babylon 5 rotates across an axis, this rotation "creates" artificial
>> gravity on the surface of the cilinder that is the station. Any object
>Right!
>> in that cilinder feels a force pushing it outwards, radially. With me
>Wrong! See next bit!
<explanation of Pedro's fundamental error snipped>
>Where am I going with this? Axis's diagram is correct, once the Star Fury
>is dropped, it has no external forces acting upon it, and must therefore
>most in a straight line from it's previous position.
<snip>
OK, so you, me and the laws of physics are so far in agreement...
>BUT! If you're in a Star Fury, you're launch bay has rotated at the same
>speed behind you, meaning that the launch bay should from a subjective
>point of view should be line up with the centre of B5's rotation.
OK, this is where we diverge. Which of our paths will Mr. Physics
follow? Let's try and find out... ;-)
If the velocity (perpendicular to the radius at launch point) is, say,
1 wibble (I like your non-SI units!), then the component of the
launchbay's velocity in that same vector = cos theta wibbles, where
theta = the angle rotated by the station since the launch at time = 0.
This function is going to vary between +1 and -1 wibbles, starting at
1 wibble. Immediately after launch, the function is going to start
dropping off from +1. Therefore, I can't agree with your statment that
the 'launch bay has rotated at the same speed behind you, meaning that
the launch bay should from a subjective point of view should be line
up with the centre of B5's rotation' as shown in your diagram below.
(This disagreement could be ASCII lack, so apologies if this is so)
>ie X POV --------> BAY | <*> core shuttle bit.
Here's a worked example. Grab a pencil and paper and things will be
clear. The figures may be different for B5, but the principle should
be the same.
Imagine the station has radius 10 wibbles.
Imagine the station takes 12 seconds to do a full rotation, i.e it
rotates at 1/12 revs per second.
The tangential speed of the Fury at launch, t=0, = 2*pi*10/12
= 5pi/3
OK, so the Fury launches at 5pi/3 wibbles/s.
Now let's move to t=3s.
The station has done a quarter turn.
The Fury has travelled 3*5pi/3 = 5pi = 15.7 wibbles (3sf)
The situation at t=3s:(x = Fury, * = center of rotation, c= cobra bay)
|<15.7>|
x------|~-,
/ |10 \
c----* |
\10 /
--___--
-------> rotation
As you can see, x-c-* isn't a straight line anymore. If you try this
with pretty much any figures, you'll get the same result, apart from
the one case in each revolution where they *are* in line.
In addition, assuming the Fury alters its pitch so that it is always
facing away from the cobra bay (which, BTW, I see no evidence of in
any of the Fury launch sequences - there's no rotation, no thruster
firing), then the aspect of the bay doors would change as they rotate
away - they would seem to become 'narrower' as the aspect angle
changes from 90 degrees to 0 degrees. There's some great shots in
'Fall of Night' and 'Severed Dreams' which fail to show this aspect
change.
And we *still* haven't explained how the Furies get out of the cobra
bays unpowered....
>I've always thought *cool* ever since I saw Star Furies launch, and still
>think it's one of the most practical launch mechanisms we seen - except
>for all the Shadow fighters launched in one go in The Long Twilight
>Struggle.
I think that's the problem. Reality substituted for what looks good.
>My brain hurts. BAD Brain... <Tvvt..>
Ditto. <S:AAB> Physics is a bitch </S:AAB>.
Take care.
//================= THIS WAS A MESSAGE FROM AXIS =================\\
// "What are you people doing standing around? /| \ / O /ŻŻŻ` \\
Hugh Robertson
Right, I'm not much cop with the maths (oops), but I'll try to explain
how I see it, fire back at will.
Ax...@netcomuk.co.uk (Axis) wrote:
>je...@cam.ac.uk (James Broughton) felt sure that it was vital for our
>survival to know that:
<James Broughtons explanation snipped>
>If you watch some of the launch sequences in which the POV of the
>observer is behind the SF for example Severed Dreams, you'll see that
>there's a good few seconds after the SF is 'dropped' during which the
>SF is unpowered, yet it somehow acquires a radial velocity away from
>the station. *Then* the thrusters kick in. This can't be correct.
>Also, given that the SF launch is unpowered, how do the SFs clear the
>launch bay in the manner shown? They should be moving *sideways* and
>then go 'BOOM!' as they are clipped by the walls of the launchbay.
The visualisation of the launch is not correct, but that doesn't
negate the launch technique. Isn't it simply a case of how far away
and how wide the launch bay doors are. And they do seem very close.
Consider:
Launchtime, Starfury is released and starts moving in a straight line
in the direction of rotation.
As soon as the fury is launched it starts moving in a straight line,
the launch bay "pulls away from it", as it continues to describe a
circle ( the launch bay is constantly being accelerated towards the
core of B5, Centripetal accelaration, the basis of the "gravity" on
B5). From the point of view of the launch bay the fury would start to
move out and back (anti-spinwards).
It isn't the fury that is moving "out" from the station, but the
launch bay moving to the side and away from the fury. The question is,
does the line of the starfury launch intersect the edge of the launch
bay doors before it gets "free" of the station? Is it possible to work
this out without figures?
>In fact, the SD shot is a perfect one to analyse. There's a shot after
>the SFs launch where they are flying, *unpowered*, directly towards
>the camera, and you can see the launch bays in the background. They
>remain directly behind the SFs in a perfect radial line.
THIS is the problem you see. They don't know exactly what's going on.
I remember from rastb5 that Gharlane used to (still does in fact) give
JMS constant (friendly - honest) abuse about science. Apparently JMS
had said that the Starfuries should spiral out from the launch bays.
On a very long pole no doubt :). And this situation (spiralling out)
is probably why they keep the furies coming out directly from the
launch bays. Wrong of course, but what the hell. It's an SF show, and
there are easier physics nits to pick.
<ASCII art snipped>
>This is definitely wrong, and it's contradicted about thirty seconds
>later when we see more SFs launch from a high external shot, and the
>launch bays revolve away from behind the SFs. I'm afraid it's all
>completely wrong. Check out the SD sequences, watch them as a
>physicist, and I'm sure you'll agree.
Yep, inconsistent.
>>>If it turns out to be the latter, I never want to here anyone slagging
>>>off S:AAB about the Hammerheads 'banking' in space again, OK? ;-)
>>OK. So S:AAB wins hands down on realism. :-)
>>Maybe not.
Not after the line like 'sun shrinking and gravity dissapearing' from
the pilot thank you very much.
>The funny thing is, if you watch closely you can see the little
>thruster clusters (great name!) fire on the Hammerheads, and they fire
>in a manner consistent with performing banks, i.e. simulating a
I watched the ep from this week, and noticed a tiny blue flare on each
wing, one up and one down, when a Hammerhead was banking. At least
they've made an effort. You've got to respect that. It's still on
probation however. Of course, they seem to have mastered artificail
gravity on their carriers, so you can't really quibble can you?
>centripetal force by varying the thrust linearly. The question is, why
>would you *want* to do that instead of using StarFury-type
>manoeuvreing?
Indeed. But then, Starfuries cannot enter atmosphere.
>Oh yeah, it looks good. Anyway the Centauri fighters (the
>crescent-shaped ones) do it too - check out the S3 title sequence.
I know they do. Eeek! When I saw this at first (Acts of Sacrifice I
think) I was a bit upset, but I'm over it now. Did the raider ships
from the first series have this behaviour too? I can't recall.
Hugh
Andrew Ansell
Axis wrote:
>
> je...@cam.ac.uk (James Broughton) felt sure that it was vital for our
> survival to know that:
>
>
> >Ok, here goes. From Fall of Night, we know that the outside of the station is
> >rotating a 60mph or about 27m/s. So when the Starfuries are lauched they would
> >have a sidewides velocity of 27m/s. However this quickly becomes insignificant
> >as even accelerating at only 4g (From Fall of Night also, we know that pilots
> >take up to 10g before blacking out), the Starfury will have a forwards(radial)
> >velocity of 39m/s after only 1 second and after 3 seconds moving fowards at
> >about 120m/s. The sideways velocity would hardly be noticable even if, as I
> >have assumed, the starfuries were not accelerating slightly off radial to
> >counteract the sideways movement.
>
> >follow a curved path (parabolic path?).
>
> Dang it! It's so hard to explain this with ASCII.
>
> I accept your explanation would be correct apart this major problem:
>
> observer is behind the SF for example Severed Dreams, you'll see that
> there's a good few seconds after the SF is 'dropped' during which the
> SF is unpowered, yet it somehow acquires a radial velocity away from
> the station. *Then* the thrusters kick in. This can't be correct.
> Also, given that the SF launch is unpowered, how do the SFs clear the
> launch bay in the manner shown? They should be moving *sideways* and
> then go 'BOOM!' as they are clipped by the walls of the launchbay.
But from a point of view inside the launch bays they would appear to
move in a straight line away for the station. The SF and the launch bay
have the same sideways velocity. Of course by the time they get a
reasonable distance from the station then they will appear to have a
sideways velocity to someone on the station. The system should work fine
as long as the launch point isn't too close to the centre of rotation of
the station. I think.
> >>If it turns out to be the latter, I never want to here anyone slagging
> >>off S:AAB about the Hammerheads 'banking' in space again, OK? ;-)
>
> >OK. So S:AAB wins hands down on realism. :-)
> >Maybe not.
>
> The funny thing is, if you watch closely you can see the little
> thruster clusters (great name!) fire on the Hammerheads, and they fire
> in a manner consistent with performing banks, i.e. simulating a
> centripetal force by varying the thrust linearly. The question is, why
> would you *want* to do that instead of using StarFury-type
> manoeuvreing?
The pilots in S:AAB are in a sitting position. This means that they
could take a 9g turn by banking but only around 4-5g by just turning. In
SF's the pilots are more in a standing/laying on their back position
(depending on which way you call up) and so the g effects are not so
important.
Andrew
Andrew Wright
In article <5c57ma$p...@news.ox.ac.uk>,
linc...@sable.ox.ac.uk.anti-spam (rupert smith) wrote:
Oh hell. It's a physics thread.
Run away! Run Away!
Andrew
--
Andrew Wright
a.wr...@max.roehampton.ac.uk <*> and...@stardock.u-net.com
http://www.personal.u-net.com/~stardock/home.html
"In cyperspace, she noticed, there were no shadows" - William Gibson
Zathras
Ax...@netcomuk.co.uk (Axis) thought long and hard on the problem,
and scribbled in his revelation:
> Zathras <zat...@mad.scientist.com> felt sure that it was vital for
> our survival to know that:
>
> >something dodgy about star fury launches, but it's not as bad as
> >Axis makes out.
this next bit sorts it out a little ;)
> OK, this is where we diverge. Which of our paths will Mr. Physics
> follow? Let's try and find out... ;-)
Think small! It's very relevant here...
<Math explanation of what's going on, agreed with - so snip!>
> Here's a worked example. Grab a pencil and paper and things will be
> clear. The figures may be different for B5, but the principle should
> be the same.
<snip - BTW I agree with the diagram etc>
This is the whole problem. The figures! A Star Fury seems to be about the
same size as your average truck, but the station is the same size as your
average village. Hence the point where the fury reaches in your diagram
represents a fairly hefty distance to travel.
I'm gonna try and punch some figures in here. We know that near the
edge of the station, the speed the inside of the station is rotating at
about 60mph. Since the Star Furies rotate at the same angular velocity,
and it looks to me (....) as if the Cobra bays are about half that
distance in. An approximation gives a speed of 30 mph as the speed at
which the Furies move about the station centre, which off course means
that they move at 30 mph when the leave the station. Now I don't know
the width of the station at that point, but if it is anywhere near say
1km or so, then this gives r as say 500m. (I'm going to approxmate that
to about third of a mile.
Given this, in your diagram, (90 degree rotation) indicates that both
the Cobra bays have travelled:
1/4 * 2 *pi * R = pi * R * .5 = (nearly) 3.14*.5*R
= .5 miles. (near enough)
This means to travel that distance at 30 mph, the fury must have been
travelling for how long? Simple arithmetic tells us 2 minutes! We *never*
see a fury staying still for that long, (even if they're not moving ala
B squared) it's thruster's always kick in after a few seconds.
It's all a matter of scale. With this level of differences of scale,
the furies would appear to move in a straight line but after about 10-
20 seconds of movement you would begin to notice the cobra bays rotating
away from you.
> As you can see, x-c-* isn't a straight line anymore. If you try this
> with pretty much any figures, you'll get the same result, apart from
> the one case in each revolution where they *are* in line.
Yep, but at the scale involved at which we see it, it's approximately
in line until the thrusters kick in - normally a couple of seconds later.
> And we *still* haven't explained how the Furies get out of the cobra
> bays unpowered....
Yep, even if JMS is on record saying the physics is right I think Ron
Thornton may have pulled the wool over JMS's eyes regarding what happens
inside the bay itself. JMS don't like math any better than the rest of
us. (I prefer compasses and rulers)
> I think that's the problem. Reality substituted for what looks good.
On the point of reality substituted for good looks, I think you might
be being a bit harsh. JMS's early GEnie posts seems to indicate a lot of
thought on the ideas we're discussing and drop launching seemed to him
to be a very utilitarian wy of doing the launch. IIRC *Somewhere* he
said that he originally thought that the Furies should be pushed out on
some kind of platform and then released (which would stop them whamming
against the station), but Ron Thornton showed him the current launch
technique, and liked what he saw.
Where does that leave us? Well, if I were Ron, I'd wonder how we could
speed up the rendering times. If, as I've shown above (convinced??) the
trajectory appears to be straight for small time periods, then it is
likely that he would have rendered as straight for that time period.
Sort of like attaching the fury to an extending arm wrt the station. This
makes the whole rotating station and Fury one hierachical model which not
only simplifies the model, but could provide a *much* faster rendering
speed. This would reduce the cost of a star fury launch in terms of CGI.
The lack of tumbling and not smacking into the cobra bays seems to an
indication that this is exactly what they've done. So it's not right,
but it is *affordable*. Result? Not one fury launched, but dozens!
OK, now where have I gone wrong, I sure I must've done somewhere - I
missed off the bit about scale last time.
Z.
----------------------------------<*>----------------------------------
"No boom today, boom tomorrow, always boom tomorrow"
Hugh Robertson
Ax...@netcomuk.co.uk (Axis) wrote:
>h...@paisley.ac.uk (Hugh Robertson) felt sure that it was vital for our
>survival to know that:
>>Right, I'm not much cop with the maths (oops), but I'll try to explain
>>how I see it, fire back at will.
>Remove thy bullet-proof vest. You seem to have the physics spot on.
That bawling physics lecturer harping on about centrifugal force
(non-existence thereof) must have done something right then. T'was
many moons ago though.
>>The visualisation of the launch is not correct, but that doesn't
>>negate the launch technique. Isn't it simply a case of how far away
>>and how wide the launch bay doors are. And they do seem very close.
><snip>
>> It isn't the fury that is moving "out" from the station, but the
>>launch bay moving to the side and away from the fury. The question is,
>>does the line of the starfury launch intersect the edge of the launch
>>bay doors before it gets "free" of the station? Is it possible to work
>>this out without figures?
>I thought about that too, but didn't want to introduce it yet lest it
>complicate matterrs (we still have to convince some people that there
>is no centripetal force acting on the Fury once the grapples release).
>But, from watching various launch sequences, I see *no* way that the
>launch bay doors are wide enough to allow the Furies to exit without
>intersection and much 'BOOM!'age. In any case, the Furies are always
>shown exiting nose-first as opposed to sideways, as would be correct.
Not sure it would be sideways, they just have to be _moving_ sideways
in relation to where the nose was pointing (radially out) at release
time.
<Snip about Gharlane & JMS >
>Really? I haven't read the rastb5 group for years. What did GoE have
>to say on this issue? He may be abusive and judgemental, but he's
>pretty strong on physics.
( I'm glad you can be so civil about Gharlane after the S:AAB
discussion)
Well, after a quick trip to dejanews here is a bit of Gharlane's
argument with the accusation.
Included message from Dejanews
---------------------------------------------------------------------------
Subject: Re: B5 gravity does not work(?)
From: ghar...@ccshp1.ccs.csus.edu (Gharlane of Eddore)
Date: 1996/06/26
Message-Id: <4qqluo$l...@news.csus.edu>
Distribution: world
References: <19960621.3...@sati.dungeon.com>
<4qnbap$m...@shelby.visix.com>
Followup-To: alt.dev.null
Summary: wrong. wrong. WRONG. also not right.
Organization: Evil Beings from Planet Eddore, Inc.
Newsgroups: rec.arts.sf.tv.babylon5.moderated
In <4qnbap$m...@shelby.visix.com> da...@visix.com (David Charlap)
writes:
>
>Actually, if you pay attention, you'll notice that the Fury's is not
>leaving the station on its own power - the engines don't fire until
>later. By pointing the ship "down", the centrifugal force (the same
>one that creates the gravity) can throw the Firy from the station.
Wrong. There is no "centrifugal force" acting on the ship, as of
the instant it is detached from the launch rails.
.....<deletia>
>
>Absolutely. The Furies would have to leave on a spiral orbit if they
>launched tangentially. Not a good idea if you want to charge out and
>enter into combat ASAP.
>
Wrong, wrong, wrong, WRONG. wrong....wrong. Also wrong.
At the moment of release, the 'Fury's vector is tangent to the
cylindrical surface defined by the motion of the launch points.
It does NOT launch on a radial line, it launches on a TANGENT.
And it does *not* "spiral* out since a 'Fury in free fall travels
in a *STRAIGHT* line.
The "Starfury launching in a spiral" schtick goes back to a few
entries perpetrated by one J. M. Straczynski, 3-4 years ago.
I don't think we ever convinced him of the invalidity of his
superstitions, but he finally just stopped talking about it.
---------------------------------------------------------------------------
There's plenty more from Gharlane if you can be bothered ploughing
through lots of accusations of stupidity on behalf of his "opposition"
(sometimes true though :-P), and some of his other posts have
calculations in them.
><re: S:AAB>
>>I watched the ep from this week, and noticed a tiny blue flare on each
>>wing, one up and one down, when a Hammerhead was banking. At least
>>they've made an effort. You've got to respect that. It's still on
>>probation however. Of course, they seem to have mastered artificial
<typo fixed >
>>gravity on their carriers, so you can't really quibble can you?
>In S:AAB canon, they have something called the Eckerley Principle
>which allows them to alter the mass of an object at will, hence FTL
>travel and artificial gravity. But I like the fact that they don't go
>in for treknobabble explanations of how the tech works. They just use
>it to make things go 'BOOM!' ;-)
Yeah, silly explanations are tiresome, irksome and a waste of screen
time. Put it all down to advances in science, and give us a decent
story.
Hugh
Axis
<snip>
>This is the whole problem. The figures! A Star Fury seems to be about the
>same size as your average truck, but the station is the same size as your
>average village. Hence the point where the fury reaches in your diagram
>represents a fairly hefty distance to travel.
I've posted a worked example with proper B5 figures elsewhere. Get
your teeth in mate!
Re: the rendering being more difficult.... Lightwave not Axis skill,
but I can see what you're driving at.
Shame JMS has defended the Furies' motion on the basis of it being
physically correct as opposed to using the 'easier to render' excuse.
Time for some R&R, methinks.
Take care.
//================= THIS WAS A MESSAGE FROM AXIS =================\\
// "What are you people doing standing around? /| \ / O /秤畔 \\
Zathras
Ax...@netcomuk.co.uk (Axis) scribbled:
> your teeth in mate!
But seriously, rather than getting my figures (60mph) from the on-screen
CGI, I based mine on Ivanova's statement in "The Fall of Night" when
Sheridan is in free fall, and hence like the furies moving independently
wrt the station - excluding the wind he'd have whipping across his face
due to turbulence, so I don't think my figure's are *that* unreliable. It's
just whether it's a realistic speed or not. (even though it is canon)
For comparison,
62m/s
*60 secs/min
=3600m/min
*60 min/hour
=216000 m/hour = 216 km/hour = 135 mph
hence your figures are larger than mine by a factor of four and the show's
by a factor of 2. Since my 30 mph comes from the show (sorta ;) I'll do
a different worked example.
You say the 'Furies don't engage thrusters until about 3 seconds in.
Rotation speed = 30mph.
My guestimate of station radius at the cobra bays of 1/3 mile.
given: (@=cobra bay)
| |
\__@__/
X-------->
Therefore in 3 secs,
Fury's position from initial point = 30miles/1200. (.025)
Cobra bay has moved 30miles/1200 about the centre of rotation.
Total distance all way round the station is .3333...*2*pi miles
which is (approx) 2.09 miles.
Angular distance from initial point is 360/(2.1/.025)
which is (360*.025)/2.1 = 9/2.1 = 4.3 degrees.
This can be viewed as fury moving .025 miles in the x direction, 0 in y.
Now, given
.center of station/rotation.
|\ (4.3 degrees) -
| \ |
| \ |
| @ | Station's radius.
+------------ Fury's tanget -
The question is does the Fury's position line up with the intersection
of the line I've drawn in? Of course not, but what's the error? Not
substantial at all (my calc doesn't do sines & cosines unfortuantely).
(I would guess since the thing's only moved 40 metres or so that the
error can't be that high all things considered...)
So if I draw | to represent the 'Fury pointing away from the station
at drop time, the diagram becomes:
.center of station/rotation.
|\ (4.3 degrees) -
| \ |
| \ |
| @ | Station's radius.
|----|---- Fury's tanget -
And hence the 'Fury *is* pointing in the wrong direction, but only by
8.6 degrees (after 3 secs) at absolute worst, 4.3 at best. Hopefully
this might put the issue to rest, as I originally stated, it is a problem
but a rotation of approximately 1.5 degrees per second is hard to spot
on something small, as would the error we're talking about here.
A moving camera to make the station appear motionless would probably
hide the effect completely.
> Shame JMS has defended the Furies' motion on the basis of it being
> physically correct as opposed to using the 'easier to render' excuse.
It was just a thought, but someone else (who you've replied to) seems
to have pointed out how the only bit I found difficult to swallow
could be handled so overall, I don't think there's anything wrong with
the physics. As for the Furies being at varying angles, I can't remember
the clip, but are you sure none of them have engaged any thrusters? If
they haven't, something's dodgy. I'll have to watch SD all over again.
The things we do. Such hardship. It's terrible....
> Time for some R&R, methinks.
... but then again, it's a good way to R&R.
That's me done. (I hope:)
Z.
----------------------------------<*>----------------------------------
"Though we are not now that strength which in days of old moved earth
and heaven, that which we are, we are; one equal temper of heroic
hearts made weak by time and fate but strong in will to strive, to seek,
to find and not to yield" -- Tennyson's "Ulysses"
Axis
h...@paisley.ac.uk (Hugh Robertson) felt sure that it was vital for our
survival to know that:
>Right, I'm not much cop with the maths (oops), but I'll try to explain
>how I see it, fire back at will.
Remove thy bullet-proof vest. You seem to have the physics spot on.
>The visualisation of the launch is not correct, but that doesn't
>negate the launch technique. Isn't it simply a case of how far away
>and how wide the launch bay doors are. And they do seem very close.
<snip>
> It isn't the fury that is moving "out" from the station, but the
>launch bay moving to the side and away from the fury. The question is,
>does the line of the starfury launch intersect the edge of the launch
>bay doors before it gets "free" of the station? Is it possible to work
>this out without figures?
I thought about that too, but didn't want to introduce it yet lest it
complicate matterrs (we still have to convince some people that there
is no centripetal force acting on the Fury once the grapples release).
But, from watching various launch sequences, I see *no* way that the
launch bay doors are wide enough to allow the Furies to exit without
intersection and much 'BOOM!'age. In any case, the Furies are always
shown exiting nose-first as opposed to sideways, as would be correct.
>THIS is the problem you see. They don't know exactly what's going on.
>I remember from rastb5 that Gharlane used to (still does in fact) give
>JMS constant (friendly - honest) abuse about science. Apparently JMS
>had said that the Starfuries should spiral out from the launch bays.
>On a very long pole no doubt :). And this situation (spiralling out)
>is probably why they keep the furies coming out directly from the
>launch bays. Wrong of course, but what the hell. It's an SF show, and
>there are easier physics nits to pick.
Really? I haven't read the rastb5 group for years. What did GoE have
to say on this issue? He may be abusive and judgemental, but he's
pretty strong on physics.
<re: S:AAB>
>I watched the ep from this week, and noticed a tiny blue flare on each
>wing, one up and one down, when a Hammerhead was banking. At least
>they've made an effort. You've got to respect that. It's still on
>probation however. Of course, they seem to have mastered artificail
>gravity on their carriers, so you can't really quibble can you?
In S:AAB canon, they have something called the Eckerley Principle
which allows them to alter the mass of an object at will, hence FTL
travel and artificial gravity. But I like the fact that they don't go
in for treknobabble explanations of how the tech works. They just use
it to make things go 'BOOM!' ;-)
Take care.
//================= THIS WAS A MESSAGE FROM AXIS =================\\
// "What are you people doing standing around? /| \ / O /ŻŻŻ` \\
Benjamin Webb
Axis wrote:
> I hate to bother you, but....
>
> The moment the grapple is released, there are NO forces acting on the
> Fury (except gravitational attraction to the station's mass - too
> small to worry about here).
>
> Call me a kipper, but I seem to recall that Newton's Laws say that, if
> the net forces acting on a body are zero, then the velocity of that
> body remains constant.
> Come on you physicists, you're being very quiet.....
Well, I'm a chemist, not a physicist, but I do know Newton's Laws
and so I have to agree with Axis here. I wouldn't go so far as to say that
centrifugal force doesn't exist, since we can only observe forces from
their effects, and in a rotating frame of reference a centrifugal force is
most certainly observed, but in an ordinary non-rotating frame of reference
(which has been used in all the arguments, I think) centrifugal force DOES
NOT exist. Centripetal force is provided by the clamps on the StarFury, and
when they disengage no forces are acting on the ship, and it flies radially
outwards. There is the problem that the continuing rotation of the station
would either clip the StarFury as it went out through the doors, but heck,
it looks pretty.
Just to give people something else to argue about, presumably the
launch has to be accurately timed. Otherwise, the StarFury squadron could
find itself heading rapidly for the zero-G docking bay...
Ben
--
Email: wadh...@sable.ox.ac.uk WWW: http://users.ox.ac.uk/~wadh0298
"Jeeves shimmered out and came back with a telegram."
- 'Carry on Jeeves (1925). Jeeves Takes Charge',
P. G. Wodehouse
Axis
ferr...@liverpool.ac.uk (Mr P.M. Martins Ferreira) felt sure that it
was vital for our survival to know that:
<Pedro and I in agreement snipped>
>So, that's my extra 2p worth. I didn't work out the numbers, anyone more
>patient than me out there...? I can tell that, if the 'gravity' in the
>bays is g, and assuming the railings' length is, say two meters (I think
>it's a safe number), the maximum speed the Fury will gain from centrifugal
>acceleration would be
> v = square root of (2*2*g) ~= 6.3 m/s
>Looking at the distance between the nose of the Fury and the Bay door, this
>would seem to work out. Fire away...
Yup. Seems right to me. I got a figure of 4.8m/s based on the fact
that the time from initial release to clearing the slide rail is, at
most, 0.5 seconds, and the radial acceleration is 1g. Of course, to be
super accurate, I should be integrating the radial acceleration over
the increasing radius of rotation, but to coin a phrase, sod it. ;-)
So, the upshot of this whole danged thread, as far as I see it is:
a) The Furies *can* clear the bay because of the 'slide effect'. If
they were simply dropped, they would likely go BOOM!
b) The Furies should be travelling *sideways* at a tangent to the
station, give or take a slight angle of arctan(4.8/62) = 4.4 degrees
c) The launch sequence, as shown for the previous 3 years, is wrong.
d) My brain hurts, I've got a bad case of tetwrist, and my phone bill
is going to be an order of magnitude larger this month.
Hopefully that should be that. The prosecution rests. ;-)
//================= THIS WAS A MESSAGE FROM AXIS =================\\
// "What are you people doing standing around? /| \ / O /秤畔 \\
Martin Hardgrave
In article <5c5cao$rsq$1...@taliesin.netcom.net.uk>, Axis
<Ax...@netcomuk.co.uk> writes
>Oh yeah, it looks good. Anyway the Centauri fighters (the
>crescent-shaped ones) do it too - check out the S3 title sequence.
Presumably Centauri and Minbari fighters incorporate some sort of
"reactionless" propulsion - Earth (and Narn?) vessels use some form of
chemical propulsion.
--
Martin
York, UK
"From hell, Hull, and Halifax, good Lord deliver us!"
Mr P.M. Martins Ferreira
Zathras (zat...@mad.scientist.com) wrote:
: This means that the *speed* (not velocity) of the ball is also 100
Erm, english not Pedro's skill... I always thought that velocity and speed
were one and the same thing, meaning, first derivative of the postion
vector?
Pedro
Mr P.M. Martins Ferreira
Rob Blake (r.p....@open.ac.uk) wrote:
Well, I am now convinced I was talking utter bollocks about the whole
centrifugal force thing. As Axis and Zathras and others so wisely pointed
out, as soon as the Fury looses contact with the station (released, that is),
this force ceases to operate (in fact, centrifugal force, as was also, very
correctly pointed out, is an artifact we use in non-inertial reference
frames (the Fury's, in this case) for convenience, but bear with me). At
least, that's the first thing that comes to mind. Unless there's some
stupid subtlety here, I'll take it as a fact. Anyway: yesterday, after
reading the replies to my initial post, I sat down, forgot about intuition,
recalled back my basic mechanics and wrote the equations. Axis and the others
are right, the Fury, IF DROPPED, should move tangentially. In the string-
stone analogy., the stone would be released tangentially, no radial
component. I keep forgetting that my intuition is lousy and only equations
work for me.
Now, having said that, I still think the Starfury's lauch works. And the
solution (?) for the conundrum comes from something I remembered yesterday,
while trying to go to sleep, and Rob pointed out:
: Well, every time I've seena starfury launch, it always seems that they
: slide down some form of runner before they exit the doors. It seems to
I think this is the key point, really. The Fury is not simply dropped,
it slides down some 'railings' for sometime, and these 'railings' lead
it almost to the bay door's entrance. While the Fury is sliding down,
it is STILL in rotation around the station, and the centrifugal force will
act upon it (after yesterday's blunder, I stand ready to be corrected),
thereby accelerating it and giving it a radial speed component that will
take him past the bay doors. Given that at the time his tangential speed
is identical to that of the bay doors - and the 'railing' carry it almost to
the bay doors - their relative motion will be zero and the Fury can
leave without any BOOM!
This alignement of the bay door with the Fury, I
would expect, would remain for sometime (as someone pointed out here, for
as long as sin(theta)~=theta, that's theta~=6 degrees), validating the
image we see of the fury leaving the bay door and remaining 'aligned'
with it. After these 6 degrees, things change, the fury's tangential speed
continues to drag him 'up' (thinking of rotation in the counter-clockwise
direction) while the station rotates, so that the bay door 'disalignes' itself
with the Fury. Rewatching Severed Dreams again, there is at least one shot
where we see this effect (and several where we see the Fury's igniting the
engines shortly after leaving the Bay's, that ruins this whole picture). There
was one of the shots that seemed a little dodgy to me, but I'm not sure it's
the physics that's wrong or the camera angle that plays tricks on the viewer).
So, that's my extra 2p worth. I didn't work out the numbers, anyone more
patient than me out there...? I can tell that, if the 'gravity' in the
bays is g, and assuming the railings' length is, say two meters (I think
it's a safe number), the maximum speed the Fury will gain from centrifugal
acceleration would be
v = square root of (2*2*g) ~= 6.3 m/s
Looking at the distance between the nose of the Fury and the Bay door, this
would seem to work out. Fire away...
: Anyone else notice this, or have I been freely hallucinating for the
: past 3 years?
Erm, you may well think that, I couldn't possibly comment upon it...
Pedro
rupert smith
> .center of station/rotation.
> |\ (4.3 degrees) -
> | \ |
> | \ |
> | @ | Station's radius.
> |----|---- Fury's tanget -
>And hence the 'Fury *is* pointing in the wrong direction, but only by
>8.6 degrees (after 3 secs) at absolute worst, 4.3 at best. Hopefully
>this might put the issue to rest, as I originally stated, it is a problem
>but a rotation of approximately 1.5 degrees per second is hard to spot
>on something small, as would the error we're talking about here.
no no no. completely and utterly wrong.
the fury still has residual angular momentum, and so will tumble at the
same rate of rotation as the station. thus, it will continue to point
away from the bay.
>That's me done. (I hope:)
>
>Z.
gas mark 5 ;-)
rupert smith
in theory, speed is a scalar and velocity is a vector. i'm not sure how
it's relevant but.....
rupert smith
In article <5ca557$g...@suns7.crosfield.co.uk>, j...@crosfield.co.uk (Jerry Cullingford) wrote:
>In article <5c9fcd$r...@news.ox.ac.uk>,
>rupert smith <linc...@sable.ox.ac.uk.anti-spam> wrote:
>>
>>the fury still has residual angular momentum, and so will tumble at the
>>same rate of rotation as the station. thus, it will continue to point
>>away from the bay.
>
>stage, if it didn't apply thrust, it'd be pointing towards the bay, which'd
>be on the other side of the station by then... but "away from the bay" is
>probably fairly close for the 33 degrees or so the station rotates in 3
>seconds.
whatever. ;-)
i suppose i should have qualified the statement, but it was 4am.
>The different vectors subsequent launches end up on is interesting though -
>
>And how do they get the furies *back* afterwards? conveyors from the zero-G
>docking bay?
yaay! back to b5! er, we're shown them flying into the main docking bay,
aren't we? we've got very little idea of what goes on inside the
station.
there's that lift/escalator that moves them into position about to be
dropped..
Ender
So what you are all trying to tell each other is that the starfuries
aren't travelling in the direction that they are pointing, when they are
dropped. They are travelling "almost sideways" out of the bay doors, but
that don't matter, cos the bay doors are also moving sideways.
Why did that ascii art just make matters worse?
Ender
--
Quote for the Week:
"Buildings Burn, People Die, but True Love is Forever"
- The Crow, John O'Barr, 1994
Ian Jefferies
Ax...@netcomuk.co.uk (Axis) wrote:
>Apologies if this was discussed way back in the dark ages of season 1.
I think it was. At one point I heard (never saw the post) that JMS thought
the Starfurys should "rotate with the station" and spiral away if they
remained unpowered. This is wrong. An explanation of this in a moment.
>Anyway, I got to Severed Dreams and I was watching the StarFury launch
>sequence. And I got to thinking about the physics involved.
>Now, as far as I can see, the Furies face away from the center of B5's
>rotation, and when they are 'dropped' their motion is along a radius
>extending from that center.
<ascii art snipped>
>Either I am missing something monumentally obvious here or there's a
>grievous error in the physics of StarFury launches. Which is it?
First some definitions:
velocity: has both magnitude and direction. Any change in the direction
implies that there is an acceleration. Acceleration implies that there is an
external force acting on the object.
speed: this is the magnitude of the velocity and is independant of the
direction in which the object is moving.
The situation is as follows:
The Starfurys starting in the COBRA bays are in the rotating frame of Babylon
5 and constained to stay there, moving in a circle. Lets call this position
in the COBRA bay point A.
When looked at from a point outside the station that is fixed in relation to
the station, but not moving relative to the station, we see the rotation of
the station and the stored Starfurys. Call this point B.
An observer at point B will see the Starfury moving at a velocity that is
continually changing. When the Starfury is release it is no longer
constrained to move with the station. However it still has the velocity
(direction and magnitude) at the time it was released. The Starfury will move
off tangentially to the station. It has no radial velocity change at the time
of launch.
As the Starfury is now free of the station it will move in a straight line
(unless acted upon by an external force). This straight line will carry it
away from the launching point A that is still rotating with the station. The
straight line and the circle touch only once so the Starfury moves away from
the station.
The view from point A on the station is a little more complicated to explain.
As the Starfury moves away from the station it will appear to drop behind the
launch position. By this I mean that it falls behind the line drawn from the
centre of B5 to point A and swept out as B5 rotates. But remember this is a
rotating frame of reference.
This can be understood by imagining a rotating disc. A point near the centre
of the disc rotates at a slower speed than a point closer to the edge, this
ensures that both points on the disc complete one revolution at the same time.
Now if you want to move from the inner point to the outer point (assuming they
are on the same radial line) then you have to accelerate through the
intervening speeds. If you move at the same speed as the inner point at all
times then clearly you will never be able to move with the outer point (it is
moving faster). If you try to move out keeping the same speed then you will
fall behind as each increase in radius requires a corresponding increase in
speed.
Phew!!! I hope that this covers the explanation Axis, it's not a trivial
thing to understand and if the Great Maker can get it wrong then at least you
are in good company :)
Demons news feed is currently over 48 hours behind, so this thread has
probably grown out of all proportion by now...
Ian.
--
E-mail: Ian.Je...@eudoxus.demon.co.uk
Author of ProgramBar, a fully featured taskbar for Win3.1, prgbr220.zip
"Quantum physicists do it discretely"
Jake S Greenland
In article: <uk7ml4t...@ASSP01.open.ac.uk> Rob Blake
<r.p....@open.ac.uk> writes:
> slide down some form of runner before they exit the doors. It seems to
> me that this would help to rotate the starfury (at least a bit) and
> also guide it out of the doors to prevent collisions.
>
> Anyone else notice this, or have I been freely hallucinating for the
> past 3 years?
This is what it has always appeared as to me as well, but then I could
just be hallucinating as well.
J
--
--------------------------------------------------------
Ja...@Mythology.demon.co.uk Sojan@Discworld
"I am not a Frog, I am a Free Womble"
--------------------------------------------------------
Mark Sinclair
In message <uk7ml4t...@ASSP01.open.ac.uk> Rob Blake wrote:
> Zathras <zat...@mad.scientist.com> writes:
> > I think to some extent that Axis is both right and wrong, there *is*
> > something dodgy about star fury launches, but it's not as bad as
> > Axis makes out.
>
> Here here. And although I agree with nearly everything you said in your
> post, there's just one other point I'd like to raise.
>
> Much was discussed here and in other articles about how the starfuries
> launch, would they clear the doors, and why don't they come out sideways.
>
> Well, every time I've seena starfury launch, it always seems that they
> slide down some form of runner before they exit the doors. It seems to
> me that this would help to rotate the starfury (at least a bit) and
> also guide it out of the doors to prevent collisions.
>
Perfectly correct. The rails would ensure that the Starfuries go out
through the doors. They would also give an extra kick of speed to
the Fury straight out of the doors. Try putting a sliding ring on
a rod and then move the rod in a circle. If you do it fast enough
the ring should go zipping off.
--
Mark Sinclair
Simon Barton
In article <uk7ml4t...@ASSP01.open.ac.uk>, Rob Blake
<r.p....@open.ac.uk> writes
>Zathras <zat...@mad.scientist.com> writes:
>> I think to some extent that Axis is both right and wrong, there *is*
>> something dodgy about star fury launches, but it's not as bad as
>> Axis makes out.
>
>Here here. And although I agree with nearly everything you said in your
>post, there's just one other point I'd like to raise.
>
> Much was discussed here and in other articles about how the starfuries
>launch, would they clear the doors, and why don't they come out sideways.
>
> Well, every time I've seena starfury launch, it always seems that they
>slide down some form of runner before they exit the doors. It seems to
>me that this would help to rotate the starfury (at least a bit) and
>also guide it out of the doors to prevent collisions.
>
>past 3 years?
>
a similar topic. There have been some things bugging me about drops, so
I thought I add my comments. My apologies If i ramble - I don't say much
in this group.
I can't see any advantage to accelerating the starfury down the drop
gear - I think they effectively slide out - as they are in a spinning
section they should drop quite nicely. If the section is at 1g and they
drop guides are 10m lone then they leave with a radial velocity of
14m/s. From earlier comments in this thread the tangential velocity is
27m/s. This doesn't cause me a problem. The starfury is not moving
freely until it leaves the rails. Even if the rails end a metre or so
inside the doors the front of the fury will be well outside when the
rear seperates. I doubt that you'd need more than a couple of metres
clearance between the 'fury and the doors if you design things
carefully.
The fact you are on a slide is the key to this in my view.
Yes I agree that you'd not get a shot of the fury moving away inline
with the drop giudes.
You can't eject the 'fury out with any force as this would affect the
station too much. If they weigh say 25 tons each and 20 are accelerated
at 4g down the rails then you cause the station to shift at something
like 1cm/s. could the beaings of the centre section take this -
especially if it's uneven and causes the centre section to wobble like a
top. If everything is weightless then you don't really need heavy
bearings between the centre section and the satationary parts. Actually,
the best arrangement is for there to be minimal contact (let both parts
"fly in formation") - actually if you don't have to pass large objects
between the two parts you would only need a simple coupling to pass
people and rings to losely keep the items together (low power magnets?).
Perhaps that's why there are two docking bays - travel between the two
parts is limited.
I digress.
Launching from the spinning part makes sense as you can launch in more
than one direction. Also the pilot's ready rooms would be in a weighted
environment, handy for keeping the troops happy. And there's that "free"
27m/s you get if you time the drop right.
Perhaps the order to drop is interpreted as "drop at the next opertune
moment" it's only 15 seconds away.
Beyond a few "artistic licence" shots I don't have any complaints with
the technique. it does seem to make sense.
Sorry if I've wandered too much and sorry that a lot of the above is
handwaving - it's a long time since my physics degree.
Think of a Polo on a drinking straw when you watch a drop happening.
Simon.
---------------------------------------------------
Simon Barton si...@sbarton.demon.co.uk
"On the stage of life someone has to work the tabs"
---------------------------------------------------
Alex Glennie
Axis <Ax...@netcomuk.co.uk.ANTISPAM-PLEASEDELETE> wrote:
>
> So, the upshot of this whole danged thread, as far as I see it is:
>
> a) The Furies *can* clear the bay because of the 'slide effect'. If
> they were simply dropped, they would likely go BOOM!
> b) The Furies should be travelling *sideways* at a tangent to the
> station, give or take a slight angle of arctan(4.8/62) = 4.4 degrees
> c) The launch sequence, as shown for the previous 3 years, is wrong.
> d) My brain hurts, I've got a bad case of tetwrist, and my phone bill
> is going to be an order of magnitude larger this month.
>
> Hopefully that should be that. The prosecution rests. ;-)
Yesss, we have a winnnneeeerrr.
<SFX> Applause, applause, crowd going nuts, etc.,etc.</SFX>
Cheers
Alex
Dang it, I've just contributed to the physics thread.
--
Physicists. Illiterates, the lot of 'em (excepting, of course, the fine
and upstanding physicists on this group (happy now?)). They have a
chance to make "Truth" and "Beauty" fundamental building blocks of the
universe, and what to do we end up with? "Top" and "Bottom". I ask you.
Axis
agle...@netcomuk.co.uk (Alex Glennie) felt sure that it was vital for
our survival to know that:
>Axis <Ax...@netcomuk.co.uk.ANTISPAM-PLEASEDELETE> wrote:
>>
>> So, the upshot of this whole danged thread, as far as I see it is:
>>
>> a) The Furies *can* clear the bay because of the 'slide effect'. If
>> they were simply dropped, they would likely go BOOM!
>> b) The Furies should be travelling *sideways* at a tangent to the
>> station, give or take a slight angle of arctan(4.8/62) = 4.4 degrees
>> c) The launch sequence, as shown for the previous 3 years, is wrong.
>> d) My brain hurts, I've got a bad case of tetwrist, and my phone bill
>> is going to be an order of magnitude larger this month.
>>
>> Hopefully that should be that. The prosecution rests. ;-)
>Yesss, we have a winnnneeeerrr.
><SFX> Applause, applause, crowd going nuts, etc.,etc.</SFX>
When truth emerges from chaos, we are *all* winners. ;-)
>Dang it, I've just contributed to the physics thread.
Bwahahahahaha! There eees nooo eescape fram ze physiks thrid!
>Physicists. Illiterates, the lot of 'em (excepting, of course, the fine
>and upstanding physicists on this group (happy now?)). They have a
>chance to make "Truth" and "Beauty" fundamental building blocks of the
>universe, and what to do we end up with? "Top" and "Bottom". I ask you.
Funny thing is, I'm an ex-physicist - my current studies reside firmly
in the social sciences and arts spheres.
But I guess once a physicist, always a physicist.
//================= THIS WAS A MESSAGE FROM AXIS =================\\
// "What are you people doing standing around? /| \ / O /ŻŻŻ` \\
rupert smith
In article <i7XbXTAD...@rauko.demon.co.uk>, Marc Read <Ma...@rauko.demon.co.uk>
<5c5h1o$uk$1...@taliesin.netcom.net.uk> <E4FAn...@liverpool.ac.uk>
<32E6E8...@mad.scientist.com> <5c6a7j$o5e$1...@taliesin.netcom.net.uk>
<32E81C...@mad.scientist.com> <5c8c1s$qdc$1...@taliesin.netcom.net.uk>
<32E838...@mad.scientist.com> <5c9fcd$r...@news.ox.ac.uk> wrote:
>(I'm no longer teaching Further Maths, but it would be a bloody good M4
>question. I also think that the dear old Coriolis Effect is somehow
>relevant, but I switched off during most of those lectures.)
i worried about it, but decided that it wasn't relevant. IIRC coriolis
only affects tangential *acceleration*
Margaret Hall
In article <32E7CD...@sable.ox.ac.uk>, Benjamin Webb
<wadh...@sable.ox.ac.uk> writes
>Axis wrote:
>> I hate to bother you, but....
>>
>> The moment the grapple is released, there are NO forces acting on the
>> Fury (except gravitational attraction to the station's mass - too
>> small to worry about here).
>>
>> Call me a kipper, but I seem to recall that Newton's Laws say that, if
>> the net forces acting on a body are zero, then the velocity of that
>> body remains constant.
>
>> Come on you physicists, you're being very quiet.....
>
> Well, I'm a chemist, not a physicist, but I do know Newton's Laws
>and so I have to agree with Axis here. I wouldn't go so far as to say that
>centrifugal force doesn't exist, since we can only observe forces from
>their effects, and in a rotating frame of reference a centrifugal force is
>most certainly observed, but in an ordinary non-rotating frame of reference
>(which has been used in all the arguments, I think) centrifugal force DOES
>NOT exist. Centripetal force is provided by the clamps on the StarFury, and
>when they disengage no forces are acting on the ship, and it flies radially
>outwards. There is the problem that the continuing rotation of the station
>would either clip the StarFury as it went out through the doors, but heck,
>it looks pretty.
>
> Just to give people something else to argue about, presumably the
>launch has to be accurately timed. Otherwise, the StarFury squadron could
>find itself heading rapidly for the zero-G docking bay...
>
> Ben
I'm not quite sure why I followed this argument with such interest. I'm
certainly no physicist (I failed A-level physics two years running far
more years ago than I care to remember :-) ). The diagrams and
mathematical formulae were fascinating though.
However, I wonder whether you haven't got the whole thing the wrong way
round? The whole problem seems to revolve (oops, pun unintentional!)
around the statement that the Star Furies just "drop" away from the
station. Now JMS is no physicist either, however, he has said several
times that he writes by watching a movie in his head over and over until
he's happy with it, then he sits down and writes what he sees.
So I take your word for it when you all point out that the maths proves
that the Star Furies can't fly out in the way they are shown if we
assume they just "drop". But what if JMS is good on seeing what happens,
but not so good on giving scientific explanations? (He often talks about
his writing as though he's a historian or reporter in the future
relaying back to us what will/might happen.) In that case it's more
likely that though the canon says the Furies are not catapulted out, the
canon is in fact wrong and there *is* some propulsive power that we are
unaware of that gives them the trajectory we observe.
In other words, I'm more prepared to trust JMS on *what* happens, rather
than the *how* and the *why*. (Especially in any scenario involving
numbers!)
Margaret
--
Margaret Hall, Gwynedd, Wales, UK
E-mail: mh...@baradel.demon.co.uk
Homepage: fantasy fiction - for magic, murder and political intrigue try
URL: http://www.baradel.demon.co.uk/
Axis
Simon Barton <si...@sbarton.demon.co.uk> felt sure that it was vital
for our survival to know that:
>In article <uk7ml4t...@ASSP01.open.ac.uk>, Rob Blake
It's hard to accurately measure the length of the slide rails from the
CGI, but they're definitely much less than 5 meters long.
In any event, the time from when the Furies start to drop to the
moment they clear the rails is at most 0.5 seconds, so the maximum
radial acceleration the Furies could experience, given 1g at the bay,
is 0.5*9.8 = 4.9m/s. You're off by a factor of at least 3.
>From earlier comments in this thread the tangential velocity is
>27m/s.
I make it 62m/s given the observed angular velocity of 1/40
revs/second and the 1g gravity at the station skin.
>This doesn't cause me a problem. The starfury is not moving
>freely until it leaves the rails. Even if the rails end a metre or so
>inside the doors the front of the fury will be well outside when the
>rear seperates. I doubt that you'd need more than a couple of metres
>clearance between the 'fury and the doors if you design things
>carefully.
>The fact you are on a slide is the key to this in my view.
>Yes I agree that you'd not get a shot of the fury moving away inline
>with the drop giudes.
Aha! This is the key point, and I'm glad you agree.
>You can't eject the 'fury out with any force as this would affect the
>station too much. If they weigh say 25 tons each and 20 are accelerated
>at 4g down the rails then you cause the station to shift at something
>like 1cm/s. could the beaings of the centre section take this -
>especially if it's uneven and causes the centre section to wobble like a
>top. If everything is weightless then you don't really need heavy
>bearings between the centre section and the satationary parts. Actually,
>the best arrangement is for there to be minimal contact (let both parts
>"fly in formation") - actually if you don't have to pass large objects
>between the two parts you would only need a simple coupling to pass
>people and rings to losely keep the items together (low power magnets?).
>Perhaps that's why there are two docking bays - travel between the two
>parts is limited.
>I digress.
Interesting point. I'd not even thought about the effect that Fury
launches would have on the station. But, given that the L5 Lagrange
point tends to equilibrium, would it matter?
>Launching from the spinning part makes sense as you can launch in more
>than one direction. Also the pilot's ready rooms would be in a weighted
>environment, handy for keeping the troops happy. And there's that "free"
>27m/s you get if you time the drop right.
>Perhaps the order to drop is interpreted as "drop at the next opertune
>moment" it's only 15 seconds away.
That's how I read it. Unfortunately for B5, they're going to go to all
that trouble of timing the drops, only to find the Furies whizzing off
in the wrong direction by about 90 degrees ;-)
>Beyond a few "artistic licence" shots I don't have any complaints with
>the technique. it does seem to make sense.
The technique is sound, the portrayal of it isn't.
>Sorry if I've wandered too much and sorry that a lot of the above is
>handwaving - it's a long time since my physics degree.
>Think of a Polo on a drinking straw when you watch a drop happening.
Apart from your 10m slide rails, you have it spot on, I think.
Congratulations sir. ;-)
Marc Read
<snip>
Good analysis. I know physics ain't a democracy, but Rupert's right, and
Axis is wrong. The only problem we have left is persuading Axis. I think
he wants a kinematical solution, not a dynamical one, folks, which is
why the arguments presented are a little off-target.
OK, Axis, at the point the grapple switches off, there are indeed no
forces acting on 'Fury. But at that moment it has a velocity vector
identical to that of the rest of the Cobra bay. Plenty of time to get
clear, especially if those rails are powered to push it out.
Hope this helps,
Marc
(I'm no longer teaching Further Maths, but it would be a bloody good M4
question. I also think that the dear old Coriolis Effect is somehow
relevant, but I switched off during most of those lectures.)
--
Marc Read
Academic -- marc...@kcl.ac.uk
Personal -- ma...@rauko.demon.co.uk http://www.rauko.demon.co.uk
"The Babylon project was our last, best, hope for peace. It failed."
Axis
Marc Read <Ma...@rauko.demon.co.uk> felt sure that it was vital for our
survival to know that:
>In article <5c9fcd$r...@news.ox.ac.uk>, rupert smith
><linc...@sable.ox.ac.uk.anti-spam> writes
><snip>
>Good analysis. I know physics ain't a democracy, but Rupert's right, and
>Axis is wrong. The only problem we have left is persuading Axis. I think
>he wants a kinematical solution, not a dynamical one, folks, which is
>why the arguments presented are a little off-target.
>OK, Axis, at the point the grapple switches off, there are indeed no
>forces acting on 'Fury. But at that moment it has a velocity vector
>identical to that of the rest of the Cobra bay. Plenty of time to get
>clear, especially if those rails are powered to push it out.
*Sigh*.
Maybe you're behind the times due to Demon lag - read the whole of the
three threads on this subject. Then watch Severed Dreams or Fall of
Night.
Check out <85421495...@eudoxus.demon.co.uk> and
<E4ICB...@liverpool.ac.uk> for two other independent confirmations
of the truth. Or talk to the two Physics PhD students I know who also
agree with me.
You're correct in that physics isn't a democracy. And in this case,
the majority is disagreeing with me - wrongly. It's just
flat-earthism, I'm afraid. The equations are plain.
I think your confusion may have arisen because my original post was
misinterpreted. From a POV inside the launch bay, the Fury *appears*
to move radially, at least while time from launch is small (in the
order of less than 5 seconds). This is fully explained elsewhere and
I'm not going to repeat it. We've also explained elsewhere how the
Furies clear the bay - by the 'slide effect'.
The point I was making was that, when viewed from an exterior POV, the
Furies should move tangentially from launch, but they don't - they're
shown to move radially. Absolutely, fundamentally, unquestionably
wrong. Sorry.
//================= THIS WAS A MESSAGE FROM AXIS =================\\
// "What are you people doing standing around? /| \ / O /秤畔 \\
John-Edward Kind
In article <5c5v75$bji$1...@taliesin.netcom.net.uk>, Axis
<Ax...@netcomuk.co.uk> writes
>ferr...@liverpool.ac.uk (Mr P.M. Martins Ferreira) felt sure that it
>was vital for our survival to know that:
>
>
>I hate to bother you, but....
>
>The moment the grapple is released, there are NO forces acting on the
>Fury (except gravitational attraction to the station's mass - too
>small to worry about here).
>
>Call me a kipper, but I seem to recall that Newton's Laws say that, if
>the net forces acting on a body are zero, then the velocity of that
>body remains constant.
>
>continue on whatever velocity vector it had at the moment the forces
>became zero, i.e. at the moment of grapple release.
What we're forgwtting is that there may be another force acting on the
fury. We don't seem to have any evidence to tell us the furies aren't
propelled forward somehow by a mechanism in the docking clamps, this
would solve all the problems about whether the furies would crash or
not, and explain the forward motion.
>
>At that moment, the velocity of the Fury is tangential, i.e.
>perpendicular to the radius of rotation. The unpowered,
>zero-force-acting-on-it Fury will continue in that tangential velocity
>until some other force acts upon it (i.e. thrusters or the launchbay
>wall WHAP!) and *not* in a perfectly radial velocity as portrayed.
>
As I said, they may be propelled clear.
>So I agree that, while the Fury is grappled to the station, it
>experiences force F = m V^2/R as you say; unfortunately that's only
>relevant when discussing the motion of a grappled Fury. As soon as
>it's cut loose, it no longer applies.
I agree it is not directly trelevent, but it is what defines the forces
acting on the fury while clamped, so effects what velocity and direction
it will have when unclamped.
>
>>I actually think that a better example is Galileo's experiment. Let a
>>ball fall from a Tower. If the Earth is rotating, why does the ball
>>fall vertically? Simple: because, at the instant of lauch, the ball has
>>the same rotation speed of Earth. Therefore, is "follows" Earth's
>>rotation precisely untill it reaches the ground. Perfect vertical. But,
>>for an outside observer, resting in relation to the Sun, for example,
>>the ball's movement would have seemed a parabol.
>
>Not convinced. It's the same principle as I outlined above. The only
>reason the ball in the tower is rotating at the same frequency as the
>Earth is that the tower is (to put it *very* crudely) a string. The
>moment you drop it from the tower (cut the string) it ceases to
>'rotate' about the Earth and is acted upon by Earth's gravity, plus
>some negligible wind resistance.
>
Yes, but as Newtons laws say, an object with no force acting on it will
carry on in the same direction at the same speed. As there is no
sideward force acting on the furies in space, they follow B5's rotation.
>To human perception, it might as well be falling straight, but it's
>really a parabola (constant velocity tangential to the Earth at the
>moment of release plus a perpendicular acceleration due to gravity).
>
>Make the tower very tall, and the parabola will be very evident. Make
>the tower tall enough, and the ball will never hit the Earth at all.
The only reason it's a parabola is because the earth's atmospher causes
friction, and the effect of the earth's gravity, causes it to slow down
(sideways, downwards it obviously accelerates.)
>>THERE! Did that do it? If not, tell me, I'll try again. And again.
>>And how many times it takes.
>
I think we're nearly there, within the next 200 years I'd say :-)
John-Edward
--
________________________________
| I'll take my time anywhere |
| Free to speak my mind anywhere |
| And I'll redefine anywhere |
| Anywhere I roam |
| Where I lay my head is home |
| |
| 'James Hetfield' |
|________________________________|
Richard Summers
John-Edward Kind wrote:
>
> In article <5c5v75$bji$1...@taliesin.netcom.net.uk>, Axis
> <Ax...@netcomuk.co.uk> writes
> >ferr...@liverpool.ac.uk (Mr P.M. Martins Ferreira) felt sure that it
> >was vital for our survival to know that:
> >
> >Fury (except gravitational attraction to the station's mass - too
> >small to worry about here).
Ahhh, but during lauch the Fury is orientated to point downwards and
it slides off the grapple by the centrifugal/pedal (whatever it's
called) force. Ever played on a round about in the park and let go by
accident only to be flung off? I think it's the same thing here.
What I would like to know, is what happens if someone jumps really
high in Bab5? Since they would be standing on the circumference as
it were and would be jump into the centre, would they move sideways
in the air?
Richard
Axis
Richard Summers <r...@mm-camb.mottmac.com> felt sure that it was vital
for our survival to know that:
>John-Edward Kind wrote:
>>
>> In article <5c5v75$bji$1...@taliesin.netcom.net.uk>, Axis
>> <Ax...@netcomuk.co.uk> writes
>> >ferr...@liverpool.ac.uk (Mr P.M. Martins Ferreira) felt sure that it
>> >was vital for our survival to know that:
>> >
>> >The moment the grapple is released, there are NO forces acting on the
>> >Fury (except gravitational attraction to the station's mass - too
>> >small to worry about here).
>Ahhh, but during lauch the Fury is orientated to point downwards and
>it slides off the grapple by the centrifugal/pedal (whatever it's
>called) force. Ever played on a round about in the park and let go by
>accident only to be flung off? I think it's the same thing here.
(For the nth time *sigh*) First, go look in a maths or physics book
and revise your rotational mechanics. OK? Now go to the park. Paint a
dot on the ground at the edge of the roundabout. Get onto the
roundabout, and get it spinning. Let go as you pass the dot. Stand
still when you land, and draw a line from where you landed to the
first dot. The line will be a tangent to the roundabout, *not* a
radius as most people seem to think.
>What I would like to know, is what happens if someone jumps really
>high in Bab5? Since they would be standing on the circumference as
>it were and would be jump into the centre, would they move sideways
>in the air?
Yes, but given the maximum thrust available by human legs and the
constraint due to the ceiling, it will be way too small to notice.
Chris Wheeler
In article <19970125....@rovanion.demon.co.uk>, Mark Sinclair
<ma...@rovanion.demon.co.uk> writes
>> > I think to some extent that Axis is both right and wrong, there *is*
>> > something dodgy about star fury launches, but it's not as bad as
>> > Axis makes out.
>>
>> Here here. And although I agree with nearly everything you said in your
>> post, there's just one other point I'd like to raise.
>>
>> Much was discussed here and in other articles about how the starfuries
>> launch, would they clear the doors, and why don't they come out sideways.
>>
>> Well, every time I've seena starfury launch, it always seems that they
>> slide down some form of runner before they exit the doors. It seems to
>> me that this would help to rotate the starfury (at least a bit) and
>> also guide it out of the doors to prevent collisions.
>>
>through the doors. They would also give an extra kick of speed to
>the Fury straight out of the doors. Try putting a sliding ring on
>a rod and then move the rod in a circle. If you do it fast enough
>the ring should go zipping off.
>
I can't work out how you would tell from the CGI, but I'd make the rails
part of a linear accelerator designed to propel the Starfuries away from
the station. This would mean that the launch mechanism and fuel would
not be carried on the Starfury, and that they would not have to engage
their own thrusters near the station.
Just letting go of the starfuries would be disasterous, as everyone has
pointed out. It seems to be standard practice to remain unpowered whilst
within the station, presumably to stop accidents and thruster damage.
Therefore, another force must be applied.
Chris
'I don't want to think about it, I don't want to talk about it,
I don't want to know about it, ... Problem Solved' - Terrorvision
Dan Hastings
Ok, I've got my head around the fact that as soon as a star fury is
released, there are no forces acting upon it and it has a velocity
at a tangent to the station. Therefore there should be no radial
speed at all, so logically it should slam into the wall, no?
Following this through, shouldn't anything anyone drops on B5 fall to
the wall instead of the floor? and then slide down the wall?
(here's betting this is one effect we _don't_ see in B5 ;)
Dan Hastings (Scatter) - dan_ha...@uk.ibm.com
Axis
John-Edward Kind <St...@vinery.demon.co.uk> felt sure that it was
vital for our survival to know that:
>What we're forgwtting is that there may be another force acting on the
>fury. We don't seem to have any evidence to tell us the furies aren't
>propelled forward somehow by a mechanism in the docking clamps, this
>would solve all the problems about whether the furies would crash or
>not, and explain the forward motion.
No propelling force exists according to, amongst other sources, JMS
himself. How the Furies clear the bay has been sorted out elsewhere.
>>>I actually think that a better example is Galileo's experiment. Let a
>>>ball fall from a Tower. If the Earth is rotating, why does the ball
>>>fall vertically? Simple: because, at the instant of lauch, the ball has
>>>the same rotation speed of Earth. Therefore, is "follows" Earth's
>>>rotation precisely untill it reaches the ground. Perfect vertical. But,
>>>for an outside observer, resting in relation to the Sun, for example,
>>>the ball's movement would have seemed a parabol.
>>
>>Not convinced. It's the same principle as I outlined above. The only
>>reason the ball in the tower is rotating at the same frequency as the
>>Earth is that the tower is (to put it *very* crudely) a string. The
>>moment you drop it from the tower (cut the string) it ceases to
>>'rotate' about the Earth and is acted upon by Earth's gravity, plus
>>some negligible wind resistance.
>>
>Yes, but as Newtons laws say, an object with no force acting on it will
>carry on in the same direction at the same speed. As there is no
>sideward force acting on the furies in space, they follow B5's rotation.
Kind of - your wording here is very woolly. In regards to the motion
of the launch bay itself, as time from launch (t) becomes larger, its
component of velocity in the direction of the Fury's motion is in the
order of cos 2pi*t/40. After t~= 9.5s, the bay is moving perpendicular
to the Fury; after t~=29.5s, the bay is moving in the exact opposite
direction to the Fury.
>>To human perception, it might as well be falling straight, but it's
>>really a parabola (constant velocity tangential to the Earth at the
>>moment of release plus a perpendicular acceleration due to gravity).
>>
>>Make the tower very tall, and the parabola will be very evident. Make
>>the tower tall enough, and the ball will never hit the Earth at all.
>The only reason it's a parabola is because the earth's atmospher causes
>friction, and the effect of the earth's gravity, causes it to slow down
>(sideways, downwards it obviously accelerates.)
No - the atmospheric friction is *one* of the reasons the ball
describes a parabola. Another, more important reason, is the fact that
the initial tangential velocity of the ball is greater than the
tangential velocity of the ground below.
Extreme case: geostationary satellites are in constant freefall, but
it so happens that because of the specific height of the orbit, the
parabola they describe as they fall is concentric to the curvature of
the Earth - hence, although they are in freefall, they never hit the
ground. So if you dropped a ball from a tower that had its top at that
geostationary level, the ball would stay in constant orbit around the
Earth. If the tower was a little taller, the ball would escape Earth's
orbit altogether. If the tower was a a little shorter, the ball would
eventually hit the Earth after orbiting it many times... that's an
*extreme* parabola. As the tower becomes progressively shorter, the
path pecomes progressively straighter - but it's always a parabola to
some degree or another, even without atmospheric friction.
>>>THERE! Did that do it? If not, tell me, I'll try again. And again.
>>>And how many times it takes.
>>
>I think we're nearly there, within the next 200 years I'd say :-)
*Some* of us are there already - others are a little further behind...
;-)
>| I'll take my time anywhere |
>| Free to speak my mind anywhere |
>| And I'll redefine anywhere |
>| Anywhere I roam |
>| Where I lay my head is home |
>| |
>| 'James Hetfield' |
Black Album! Pthhhhrrrbld!
'Do you see fear what I fear?
Living properly -
Truths to you are lies to me.'
James Hetfield *before* he sold out.
Marc Read
In article <5cfn03$9b2$1...@taliesin.netcom.net.uk>, Axis
<Ax...@netcomuk.co.uk.ANTISPAM-PLEASEDELETE> writes
<snip>
>Maybe you're behind the times due to Demon lag - read the whole of the
>three threads on this subject. Then watch Severed Dreams or Fall of
>Night.
Yes, major Demon lag. I'm still arguing about the point you mention
below. I think I'll sit back and contribute only to slow-moving threads
until demon gets it together, :)
>
>Check out <85421495...@eudoxus.demon.co.uk> and
><E4ICB...@liverpool.ac.uk> for two other independent confirmations
>of the truth. Or talk to the two Physics PhD students I know who also
>agree with me.
Hrm. I know a couple of DPhil students who don't... doesn't prove
anything. In "Surely you're joking, Mister Feynmann" Richard F gives a
"simple" physics problem which physicists always think is completely
trivial. The problem is that half of them say one thing, half another.
And IIRC it's about rotation....
<snip>
>
>I think your confusion may have arisen because my original post was
>misinterpreted. From a POV inside the launch bay, the Fury *appears*
>to move radially, at least while time from launch is small (in the
>order of less than 5 seconds). This is fully explained elsewhere and
>I'm not going to repeat it. We've also explained elsewhere how the
>Furies clear the bay - by the 'slide effect'.
>
Point taken. This is what I was arguing about.
>The point I was making was that, when viewed from an exterior POV, the
>Furies should move tangentially from launch, but they don't - they're
>shown to move radially. Absolutely, fundamentally, unquestionably
>wrong. Sorry.
>
I have two differing intuitions (aagh!). I see your argument. Can
someone please let me know what has happened to all that nice angular
momentum the 'fury has, though? Then I'll be happy.
Marc
FWIW, Axis, I'm also an ex-physics student who's now in the humanities.
Trouble is, I also teach (part-time) physics to 'A'-level. Luckily, the
a-level is now so easy they wouldn't dream of asking things like that.
<sigh of relief>
</worried thought>
If anyone here works for an exam board -- DON'T DO IT!
</wt>
Richard Summers
>
> (For the nth time *sigh*) First, go look in a maths or physics book
> and revise your rotational mechanics. OK? Now go to the park. Paint a
> dot on the ground at the edge of the roundabout. Get onto the
> roundabout, and get it spinning. Let go as you pass the dot. Stand
> still when you land, and draw a line from where you landed to the
> first dot. The line will be a tangent to the roundabout, *not* a
> radius as most people seem to think.
Oh dear, whats the matter, am I getting you down? Actually I
hated rotation mechanics, prefered optics instead. But pleeease
don't make me look at a physics book again, I have only just
got over my recurring nightmare of sitting an exam without
any revision and wearing no clothes! ( I'm not joking ).
Back to the discussion. Yes of course you are correct. Let me
try again, Bab5 has a very large diameter. The Star Fury is
located on two prongs (like a fork lift) which are orientated
to point the Star Fury outwards. These prongs also stop the
tangental movement. So when the grapple is released the Star
Fury will remain where it is until, the station has rotated
a bit. Then there will be a small velocity component in the
direction of the grapple prongs, so the Star Fury starts moving
radially. The more the station rotates larger the radial
acceleration.
mmmm come to think of it, I'm now not convinced that the above
is going to provide enough radial velocity. If the radial speed
is great enough then the lauch should be ok.
> >What I would like to know, is what happens if someone jumps really
> >high in Bab5? Since they would be standing on the circumference as
> >it were and would be jump into the centre, would they move sideways
> >in the air?
>Yes, but given the maximum thrust available by human legs and the
>constraint due to the ceiling, it will be way too small to notice.
I wasn't being that serious (;-} ( I have a beard )
Richard
r...@mm-camb.mottmac.com
Axis
agle...@netcomuk.co.uk (Alex Glennie) felt sure that it was vital for
our survival to know that:
(re: Feynman's problem)
>P.S. I guess it's possible you might need a huge super-duper computer to
>actually _prove_ the right answer
>--
Or, on a more practical note, a garden sprayer, a tub of water and an
Aquavac... ;-)
//================= THIS WAS A MESSAGE FROM AXIS =================\\
// "What are you people doing standing around? /| \ / O /ŻŻŻ` \\
rupert smith
In article <1997012819...@dialup-11-28.netcomuk.co.uk>, agle...@netcomuk.co.uk (Alex Glennie) wrote:
>> >Imagine a rotating garden sprayer (you know, the things you stick in the
>> >middle of your lawn and connect a hose to) underwater. Pump water
>> >through and it spins just like it would on dry land. Now *suck* water
>> >back through it. Does it spin in the same direction or the opposite
>> >direction?
>In the book "Genius" by James Gleick, an excellent biography of the
>great man, Feynman's answer is given on page ....well, perhaps I'll
>leave it as an exercise for the reader to work out. But there is an
>absolutly categorical answer. Same or opposite? Opposite or same? You
>decide.
lies! lies! lies!
it doesn't move at all.
think angular momentum: the water going up the central pipe doesn't have
much, so the rotor won't.
>P.S. I guess it's possible you might need a huge super-duper computer to
>actually _prove_ the right answer
alternatively, a basin of water and a bent straw. experimental physics
rules!
Neale Grant
In article <OzbQVAA1...@rauko.demon.co.uk>,
Marc Read <Ma...@rauko.demon.co.uk> wrote:
>In article <5cfn03$9b2$1...@taliesin.netcom.net.uk>, Axis
><Ax...@netcomuk.co.uk.ANTISPAM-PLEASEDELETE> writes
>><E4ICB...@liverpool.ac.uk> for two other independent confirmations
>>of the truth. Or talk to the two Physics PhD students I know who also
>>agree with me.
>Hrm. I know a couple of DPhil students who don't... doesn't prove
>anything. In "Surely you're joking, Mister Feynmann" Richard F gives a
>"simple" physics problem which physicists always think is completely
>trivial. The problem is that half of them say one thing, half another.
>And IIRC it's about rotation....
Ah, if it's the one I'm thinking of...
If you thought the StarFury argument was bitter, Feynman's one could run
and run (if anyone is foolish enough to take it up):
Imagine a rotating garden sprayer (you know, the things you stick in the
middle of your lawn and connect a hose to) underwater. Pump water
through and it spins just like it would on dry land. Now *suck* water
back through it. Does it spin in the same direction or the opposite
direction?
(I don't pretend to know the answer.)
Neale
--
Brit(S) H++:U--:- a20 s+++:- hf+>++ b! m! y! X---: P-- S++ M-- R--->$
!A+++ C- T! TV- Ci- MuR+++ACIPZ+ Am+++ Ac? B+>+++ V-- (v1.1)
http://users.ox.ac.uk/~univ0155/ - Home of TQE and ^^^the BritCode^^^
AJ Groos
Your diagrams are correct.
The ship would proceed tangentially from the point of release.
It would go at a straight line vector.
For a time the ship would look like it was following the path of the
hole it dropped out of but after a few degrees of arc they would no
longer look aligned.
Ax...@netcomuk.co.uk (Axis) wrote:
>je...@cam.ac.uk (James Broughton) felt sure that it was vital for our
>survival to know that:
><snip>
>>Ok, here goes. From Fall of Night, we know that the outside of the station is
>>rotating a 60mph or about 27m/s. So when the Starfuries are lauched they would
>>have a sidewides velocity of 27m/s. However this quickly becomes insignificant
>>as even accelerating at only 4g (From Fall of Night also, we know that pilots
>>take up to 10g before blacking out), the Starfury will have a forwards(radial)
>>velocity of 39m/s after only 1 second and after 3 seconds moving fowards at
>>about 120m/s. The sideways velocity would hardly be noticable even if, as I
>>have assumed, the starfuries were not accelerating slightly off radial to
>>counteract the sideways movement.
>>All in all they would appear to come straight out although they would actually
>>follow a curved path (parabolic path?).
>Dang it! It's so hard to explain this with ASCII.
>I accept your explanation would be correct apart this major problem:
>If you watch some of the launch sequences in which the POV of the
>observer is behind the SF for example Severed Dreams, you'll see that
>there's a good few seconds after the SF is 'dropped' during which the
>SF is unpowered, yet it somehow acquires a radial velocity away from
>the station. *Then* the thrusters kick in. This can't be correct.
>Also, given that the SF launch is unpowered, how do the SFs clear the
>launch bay in the manner shown? They should be moving *sideways* and
>then go 'BOOM!' as they are clipped by the walls of the launchbay.
>In fact, the SD shot is a perfect one to analyse. There's a shot after
>the SFs launch where they are flying, *unpowered*, directly towards
>the camera, and you can see the launch bays in the background. They
>remain directly behind the SFs in a perfect radial line.
>More crappy ASCII art:
> \
> \ Launchbay remains 'behind' SF instead of rotating away
> oooo
>center o XX,, Vector of unpowered SF POV of observer
>* of o XX> -----> <-- o>
>rot. o XX``
> oooo
> / | Launchbay should be moving in this direction
> / |
> v
>This is definitely wrong, and it's contradicted about thirty seconds
>later when we see more SFs launch from a high external shot, and the
>launch bays revolve away from behind the SFs. I'm afraid it's all
>completely wrong. Check out the SD sequences, watch them as a
>physicist, and I'm sure you'll agree.
>Oh well, at least it's on-topic. ;-)
>>>If it turns out to be the latter, I never want to here anyone slagging
>>>off S:AAB about the Hammerheads 'banking' in space again, OK? ;-)
>>OK. So S:AAB wins hands down on realism. :-)
>>Maybe not.
>The funny thing is, if you watch closely you can see the little
>thruster clusters (great name!) fire on the Hammerheads, and they fire
>in a manner consistent with performing banks, i.e. simulating a
>centripetal force by varying the thrust linearly. The question is, why
>would you *want* to do that instead of using StarFury-type
>manoeuvreing?
>Oh yeah, it looks good. Anyway the Centauri fighters (the
>crescent-shaped ones) do it too - check out the S3 title sequence.
> //================= THIS WAS A MESSAGE FROM AXIS =================\\
> // "What are you people doing standing around? /| \ / O /秤畔 \\
Andy Lupini
[snip]
> I have two differing intuitions (aagh!). I see your argument. Can
> someone please let me know what has happened to all that nice angular
> momentum the 'fury has, though? Then I'll be happy.
>
>
> Marc
>
> FWIW, Axis, I'm also an ex-physics student who's now in the humanities.
> Trouble is, I also teach (part-time) physics to 'A'-level. Luckily, the
> a-level is now so easy they wouldn't dream of asking things like that.
> <sigh of relief>
Aarrggh! I wan't going to post on this rather uninteresting thread
until I read this! How are you supposed to be able to teach physics if
you can't do this?
Right in your a-level terms- angular momentum is:
J=mvr where the symbols have their usual meanings, other than r which is...
perpendicular distance ( to v ).
Right, you seem to have grasped the concept that it will fly off at a
tangent, so in space given that v and now r this perpendicular
distance are assumed constant, what do you _think_ has happened to
this angular momentum?
(please please please don't say- aha m must change!)
> </worried thought>
> If anyone here works for an exam board -- DON'T DO IT!
> </wt>
Hehehehe, now that is an idea:
Assuming a cylindrical space-station...
Andy
Jon Turner
Ax...@netcomuk.co.uk.ANTISPAM-PLEASEDELETE (Axis) dribbled incoherently
:
>Marc Read <Ma...@rauko.demon.co.uk> felt sure that it was vital for our
>survival to know that:
>
>>In article <5c9fcd$r...@news.ox.ac.uk>, rupert smith
>><linc...@sable.ox.ac.uk.anti-spam> writes
>
>><snip>
>
>>Good analysis. I know physics ain't a democracy, but Rupert's right, and
>>Axis is wrong. The only problem we have left is persuading Axis. I think
>>he wants a kinematical solution, not a dynamical one, folks, which is
>>why the arguments presented are a little off-target.
>
>>OK, Axis, at the point the grapple switches off, there are indeed no
>>forces acting on 'Fury. But at that moment it has a velocity vector
>>identical to that of the rest of the Cobra bay. Plenty of time to get
>>clear, especially if those rails are powered to push it out.
>
>*Sigh*.
>
>Maybe you're behind the times due to Demon lag - read the whole of the
>three threads on this subject. Then watch Severed Dreams or Fall of
>Night.
>
>Check out <85421495...@eudoxus.demon.co.uk> and
><E4ICB...@liverpool.ac.uk> for two other independent confirmations
>of the truth. Or talk to the two Physics PhD students I know who also
>agree with me.
>
>You're correct in that physics isn't a democracy. And in this case,
>the majority is disagreeing with me - wrongly. It's just
>flat-earthism, I'm afraid. The equations are plain.
>
>I think your confusion may have arisen because my original post was
>misinterpreted. From a POV inside the launch bay, the Fury *appears*
>to move radially, at least while time from launch is small (in the
>order of less than 5 seconds). This is fully explained elsewhere and
>I'm not going to repeat it. We've also explained elsewhere how the
>Furies clear the bay - by the 'slide effect'.
>
>The point I was making was that, when viewed from an exterior POV, the
>Furies should move tangentially from launch, but they don't - they're
>shown to move radially. Absolutely, fundamentally, unquestionably
>wrong. Sorry.
>
Correct and well sumerised.
but hey what about sheridan..
<duck and run>.
Jon Turner J...@acme-labs.demon.co.uk
--
"The sea, Pinky.
'Roll on thou deep and dark blue ocean, Roll',Byron"
"Good one Brain. Hey I got one.
'Hey you kids stay out of the rip tide.',Baywatch"
Alex Glennie
Axis <Ax...@netcomuk.co.uk.ANTISPAM-PLEASEDELETE> wrote:
> univ...@sable.ox.ac.uk (Neale Grant) felt sure that it was vital for
> our survival to know that:
>
> >If you thought the StarFury argument was bitter, Feynman's one could run
> >and run (if anyone is foolish enough to take it up):
>
> >Imagine a rotating garden sprayer (you know, the things you stick in the
> >middle of your lawn and connect a hose to) underwater. Pump water
> >through and it spins just like it would on dry land. Now *suck* water
> >back through it. Does it spin in the same direction or the opposite
> >direction?
>
> Ah, that old chestnut. It's not at all a trivial problem because it
> involves fluid dynamics which always complicates matters far more than
> you might at first expect.
> Without some hardcore supercomputers I wouldn't even attempt to model
> such a problem.
<Lee and Herring> Ahhhhhhh </Lee and Herring> But there _is_ a simple
answer. The trouble is, you need a sense of physical intuition pretty
much as good as Feynmans to work it out. You do _not_ need any numbers.
In the book "Genius" by James Gleick, an excellent biography of the
great man, Feynman's answer is given on page ....well, perhaps I'll
leave it as an exercise for the reader to work out. But there is an
absolutly categorical answer. Same or opposite? Opposite or same? You
decide.
Look, honestly, I've got _nothing_ against physicists.
Cheers
Alex
P.S. I guess it's possible you might need a huge super-duper computer to
actually _prove_ the right answer
--
.sig undergoing vetting to ensure no offence is inadvertently caused to
anyone, honest, guv, last thing I'd want...
Axis
j...@acme-labs.demon.co.uk (Jon Turner) felt sure that it was vital for
our survival to know that:
<snip>
>>The point I was making was that, when viewed from an exterior POV, the
>>Furies should move tangentially from launch, but they don't - they're
>>shown to move radially. Absolutely, fundamentally, unquestionably
>>wrong. Sorry.
>>
>Correct and well sumerised.
>but hey what about sheridan..
><duck and run>.
Depends... if the pit was deep enough, he'd slam into the walls
first... not necessarily a *bad* thing, mind you ;-)
Axis
Marc Read <Ma...@rauko.demon.co.uk> felt sure that it was vital for our
survival to know that:
>In article <5cfn03$9b2$1...@taliesin.netcom.net.uk>, Axis
><Ax...@netcomuk.co.uk.ANTISPAM-PLEASEDELETE> writes
(snip)
>>The point I was making was that, when viewed from an exterior POV, the
>>Furies should move tangentially from launch, but they don't - they're
>>shown to move radially. Absolutely, fundamentally, unquestionably
>>wrong. Sorry.
>>
>I have two differing intuitions (aagh!). I see your argument. Can
>someone please let me know what has happened to all that nice angular
>momentum the 'fury has, though? Then I'll be happy.
This has been accounted for. The Fury conserves its momentum, just as
it should do, so it is rotating at pi/20 rad/s.
>FWIW, Axis, I'm also an ex-physics student who's now in the humanities.
>Trouble is, I also teach (part-time) physics to 'A'-level. Luckily, the
>a-level is now so easy they wouldn't dream of asking things like that.
><sigh of relief>
></worried thought>
>If anyone here works for an exam board -- DON'T DO IT!
></wt>
If you think physics is a bitch, you should try studying Mandarin
Chinese... I have just got back from two of my Chinese exams and this
thread seems a breeze in comparison.
//================= THIS WAS A MESSAGE FROM AXIS =================\\
// "What are you people doing standing around? /| \ / O /ŻŻŻ` \\
Axis
univ...@sable.ox.ac.uk (Neale Grant) felt sure that it was vital for
our survival to know that:
>If you thought the StarFury argument was bitter, Feynman's one could run
>and run (if anyone is foolish enough to take it up):
>Imagine a rotating garden sprayer (you know, the things you stick in the
>middle of your lawn and connect a hose to) underwater. Pump water
>through and it spins just like it would on dry land. Now *suck* water
>back through it. Does it spin in the same direction or the opposite
>direction?
Ah, that old chestnut. It's not at all a trivial problem because it
involves fluid dynamics which always complicates matters far more than
you might at first expect.
Without some hardcore supercomputers I wouldn't even attempt to model
such a problem.
>(I don't pretend to know the answer.)
Me neither, but I'd *like* to know so long as it involved zero
intellectual effort to do so - my mind is addled enough from today's
Chinese exams ;-)
//================= THIS WAS A MESSAGE FROM AXIS =================\\
// "What are you people doing standing around? /| \ / O /秤畔 \\
rupert smith
In article <32f6800a...@news.u-net.com>, mhub...@hulmecc.u-net.com (Mike Hubbard) wrote:
>On Tue, 28 Jan 97 20:48:39 GMT, linc...@sable.ox.ac.uk.anti-spam
>(rupert smith) wrote:
>
>>In article <1997012819...@dialup-11-28.netcomuk.co.uk>,
> agle...@netcomuk.co.uk (Alex Glennie) wrote:
>
>>>P.S. I guess it's possible you might need a huge super-duper computer to
>>>actually _prove_ the right answer
>>
>>alternatively, a basin of water and a bent straw. experimental physics
>>rules!
>>
>That's cheating! How are the universities going to get funding for their
>supercomputers now? You can't make the students go back to the PC
>version of DOOM after they have played it on a Cray.
i've played doom on a sparcstation: i wasn't impressed.
>Or is that what I am supposed to think?
we know what you are thinking!
(and you should know better..)
Marc Read
In article <5ckj6o$e...@lyra.csx.cam.ac.uk>, Andy Lupini <arl1000@talia.g
irton.cam.ac.uk> writes
<snip>
>
> Aarrggh! I wan't going to post on this rather uninteresting thread
>until I read this! How are you supposed to be able to teach physics if
>you can't do this?
>
I'll have to check when I posted. I must have been drunk or hungover. Of
course, you're quite correct, and I'm duly ashamed of myself. It's
exactly the sort of thing I was teaching last year. Mea culpa. I was
unduly influenced by recently having to teach the General Circulation of
the Atmosphere (Coriolis stuff) which is a different kettle of bananas.
(As far as teaching physics goes, as I say -- this sort of thing is WAY
beyond the syllabus. And though it would be nice to have some bright
pupils who didn't switch off when we do something which isn't printed in
the syllabus booklet, it's not a situation I find myself in...)
> Right in your a-level terms- angular momentum is:
>
>J=mvr where the symbols have their usual meanings, other than r which is...
>perpendicular distance ( to v ).
>
> Right, you seem to have grasped the concept that it will fly off at a
>tangent, so in space given that v and now r this perpendicular
>distance are assumed constant, what do you _think_ has happened to
>this angular momentum?
>
> (please please please don't say- aha m must change!)
>
As I say, I really can't think why I posted it. After all those bloody
further maths questions about cricket balls hitting hinged posts, and
things thrown from turntables....
Mike Hubbard
On Tue, 28 Jan 97 20:48:39 GMT, linc...@sable.ox.ac.uk.anti-spam
(rupert smith) wrote:
>In article <1997012819...@dialup-11-28.netcomuk.co.uk>, agle...@netcomuk.co.uk (Alex Glennie) wrote:
(Underwater garden sprinkler set to suck snipped)
>lies! lies! lies!
>it doesn't move at all.
>
>think angular momentum: the water going up the central pipe doesn't have
>much, so the rotor won't.
>
>>P.S. I guess it's possible you might need a huge super-duper computer to
>>actually _prove_ the right answer
>
>alternatively, a basin of water and a bent straw. experimental physics
>rules!
>
That's cheating! How are the universities going to get funding for their
supercomputers now? You can't make the students go back to the PC
version of DOOM after they have played it on a Cray.
If blowing water out of the sprinkler makes it go clockwise, then
sucking it in will make it go anti-clockwise, won't it? If the rotors
are at 12 O'clock and 6 O'clock, pointing towards 11 and 5, then that is
where the water will be blown to or sucked from.
Just like sucking or blowing on a toy windmill.
Or is that what I am supposed to think?
--
<*> Mike Hubbard. Hulme Community Computing. 0161-226 5088
"That is SO not my motto". "Hasta la vista, virtual teapot".
Simon Barton
In article <5cfnq1$9oj$1...@taliesin.netcom.net.uk>, Axis <Axis@netcomuk.
co.uk.ANTISPAM-PLEASEDELETE> writes
>Simon Barton <si...@sbarton.demon.co.uk> felt sure that it was vital
>for our survival to know that:
>
>><r.p....@open.ac.uk> writes
>>>Zathras <zat...@mad.scientist.com> writes:
snipping freely....
>>I can't see any advantage to accelerating the starfury down the drop
>>gear - I think they effectively slide out - as they are in a spinning
>>section they should drop quite nicely. If the section is at 1g and they
>>drop guides are 10m lone then they leave with a radial velocity of
>>14m/s.
>
>It's hard to accurately measure the length of the slide rails from the
>CGI, but they're definitely much less than 5 meters long.
>
>In any event, the time from when the Furies start to drop to the
>moment they clear the rails is at most 0.5 seconds, so the maximum
>radial acceleration the Furies could experience, given 1g at the bay,
>is 0.5*9.8 = 4.9m/s. You're off by a factor of at least 3.
>
>>From earlier comments in this thread the tangential velocity is
>>27m/s.
>
>I make it 62m/s given the observed angular velocity of 1/40
>revs/second and the 1g gravity at the station skin.
>
Well, it was a handwaving sort of argument. Doesn't matter much if the
rails are inline with the doors. Do we ever see a launch where they swing
the furies into position before opening the doors?
>>This doesn't cause me a problem. The starfury is not moving
>>freely until it leaves the rails. Even if the rails end a metre or so
>>inside the doors the front of the fury will be well outside when the
>>rear seperates. I doubt that you'd need more than a couple of metres
>>clearance between the 'fury and the doors if you design things
>>carefully.
>>The fact you are on a slide is the key to this in my view.
>
>>Yes I agree that you'd not get a shot of the fury moving away inline
>>with the drop giudes.
>
>Aha! This is the key point, and I'm glad you agree.
>
Agree on the slide or the lack of the view?
>>You can't eject the 'fury out with any force as this would affect the
>>station too much. If they weigh say 25 tons each and 20 are accelerated
>>at 4g down the rails then you cause the station to shift at something
>>like 1cm/s. could the beaings of the centre section take this -
>>especially if it's uneven and causes the centre section to wobble like a
>>top. If everything is weightless then you don't really need heavy
>>bearings between the centre section and the satationary parts. Actually,
>>the best arrangement is for there to be minimal contact (let both parts
>>"fly in formation") - actually if you don't have to pass large objects
>>between the two parts you would only need a simple coupling to pass
>>people and rings to losely keep the items together (low power magnets?).
>>Perhaps that's why there are two docking bays - travel between the two
>>parts is limited.
>>I digress.
>
>Interesting point. I'd not even thought about the effect that Fury
>launches would have on the station. But, given that the L5 Lagrange
>point tends to equilibrium, would it matter?
>
Not up on the locations and stabilities of the Lagrange points - if the local
planet has no moon, and we see the station both in sunlight and in
shadow, then is it really at a stable point? Can someone clarify this for
me?
The problem is not the whole station drifting, but the rotating and non-
rotating parts drifting relative to each other as this will place potentialy
nasty loads on the bearings.
>>Launching from the spinning part makes sense as you can launch in more
>>than one direction. Also the pilot's ready rooms would be in a weighted
>>environment, handy for keeping the troops happy. And there's that "free"
>>27m/s you get if you time the drop right.
>>Perhaps the order to drop is interpreted as "drop at the next opertune
>>moment" it's only 15 seconds away.
>
>That's how I read it. Unfortunately for B5, they're going to go to all
>that trouble of timing the drops, only to find the Furies whizzing off
>in the wrong direction by about 90 degrees ;-)
>
The launch 90 degrees out. They've had the chance for quite a bit of
practice. The crunch is gaining the free velocity, which would be rather
handy.
(at the expense of the momentum of the rotating section?)
Would the delay from an instant launch be preferable to the huge velocity
change to end up going the right way?
>>Beyond a few "artistic licence" shots I don't have any complaints with
>>the technique. it does seem to make sense.
>
>The technique is sound, the portrayal of it isn't.
>
Hear here. But it looks rather nice.
>>Sorry if I've wandered too much and sorry that a lot of the above is
>>handwaving - it's a long time since my physics degree.
>
>>Think of a Polo on a drinking straw when you watch a drop happening.
>
>Apart from your 10m slide rails, you have it spot on, I think.
>Congratulations sir. ;-)
>
Thankyou.
Simon
---------------------------------------------------
Simon Barton si...@sbarton.demon.co.uk
"On the stage of life someone has to work the tabs"
---------------------------------------------------
Andy Lupini
In article <85464806...@eudoxus.demon.co.uk>,
Ian.Je...@eudoxus.demon.co.uk (Ian Jefferies) writes:
> arl...@talia.girton.cam.ac.uk (Andy Lupini) wrote:
>
> There should be no change in the rotational rate of the station because of the
> Startfury launch: the total amount of energy is conserved.
Yes.
> Angular momentum is defined as the velocity multiplied by the closest distance
> that the straight line path passes to the centre of rotation. The item
> doesn't have to be attached to be a part of the calculation.
No, for 2 reasons, can you guess what they are?
Andy
mag...@aol.com
In article <32E7CD...@sable.ox.ac.uk>, Benjamin Webb
<wadh...@sable.ox.ac.uk> writes:
> I wouldn't go so far as to say that
>centrifugal force doesn't exist, since we can only observe forces from
>their effects, and in a rotating frame of reference a centrifugal force
is
>most certainly observed
Errrmm. No it`s not. Think of when you are in a car going round a bend. At
first you appear move outwards from the bend, you are in actuall fact
moving at a tangent to bend. This is not because you are accelerating
(providing the car does not accelerate and the bend remains constant)
outwards, moreover it is that your mommentum from earlier maintains your
motion for some time in that direction.
No acceleration outwards so far, therefore no force.
Next you hit the side of the car and think that you exert a force on it.
Not so. You feel the reaction force from the door, but this acts towards
the centre of revolution and is a manifestation of the cars acceleration
towards the centre. Thus a force acts on you towards the centre but you
never actually exert a force on the door.
If you did you would not accelerate around the bend at the same rate as
the car and would find yourself moving backwards relative to the car,
which does not happen.
SO to reiterate. Centrifugal force exists neither from the standpoint of
an outside viewer nor from someone inside the system.
I just know someone is reading this who knows more than me.....
Mag...@aol.com
Alex Glennie
rupert smith <linc...@sable.ox.ac.uk.anti-spam> wrote:
> In article <1997012819...@dialup-11-28.netcomuk.co.uk>,
>agle...@netcomuk.co.uk (Alex Glennie) wrote:
> >In the book "Genius" by James Gleick, an excellent biography of the
> >great man, Feynman's answer is given on page ....well, perhaps I'll
> >leave it as an exercise for the reader to work out. But there is an
> >absolutly categorical answer. Same or opposite? Opposite or same? You
> >decide.
>
> lies! lies! lies!
> it doesn't move at all.
Darn it, you're not allowed to get the answer straight away. Where's the
fun in that? (In case anybody's interested, the answer is on page 107
of said book, in my paperback)
> think angular momentum: the water going up the central pipe doesn't have
> much, so the rotor won't.
>
> >P.S. I guess it's possible you might need a huge super-duper computer to
> >actually _prove_ the right answer
>
> alternatively, a basin of water and a bent straw. experimental physics
> rules!
The great thing is that in "Surely You're Joking..", Feynman actually
does do the experiment, and gets the correct answer (i.e it doesn't
move). But the fact that the experiment then explodes distracts
everybody, cue arguments.
Cheers
Alex
--
"OK, OK, let me get this right ... if I wanna be your lover, I've gotta
get with your friends? What the hell are you *talking* about"
(blatantly ripped off from somebody, sorry)
Ian Jefferies
arl...@talia.girton.cam.ac.uk (Andy Lupini) wrote:
> Right in your a-level terms- angular momentum is:
>J=mvr where the symbols have their usual meanings, other than r which is...
>perpendicular distance ( to v ).
> Right, you seem to have grasped the concept that it will fly off at a
>tangent, so in space given that v and now r this perpendicular
>distance are assumed constant, what do you _think_ has happened to
>this angular momentum?
> (please please please don't say- aha m must change!)
There should be no change in the rotational rate of the station because of the
Startfury launch: the total amount of energy is conserved.
Angular momentum is defined as the velocity multiplied by the closest distance
that the straight line path passes to the centre of rotation. The item
doesn't have to be attached to be a part of the calculation.
Ian.
--
E-mail: Ian.Je...@eudoxus.demon.co.uk
Author of ProgramBar, a fully featured taskbar for Win3.1, prgbr220.zip
"Quantum physicists do it discretely"
Martin Hardgrave
In article <32f6800a...@news.u-net.com>, Mike Hubbard
<mhub...@hulmecc.u-net.com> writes
>That's cheating! How are the universities going to get funding for their
>supercomputers now? You can't make the students go back to the PC
>version of DOOM after they have played it on a Cray.
>
>If blowing water out of the sprinkler makes it go clockwise, then
>sucking it in will make it go anti-clockwise, won't it? If the rotors
>are at 12 O'clock and 6 O'clock, pointing towards 11 and 5, then that is
>where the water will be blown to or sucked from.
>
>Just like sucking or blowing on a toy windmill.
>
>Or is that what I am supposed to think?
Probably. I'd hazard a guess that it:
a continues to rotate the same way, as the water in the nozzle has
some momentum which is transferred to the end of the head before going
up the main tube, or
b rotates the opposite way due to the moving liquid in front of
the nozzle having a lower pressure which tends to "suck" the nozzle
round, or
c doesn't rotate at all.
--
Martin
York, UK
"From hell, Hull, and Halifax, good Lord deliver us!"
Benjamin Webb
mag...@aol.com wrote:
>
> In article <32E7CD...@sable.ox.ac.uk>, Benjamin Webb
> <wadh...@sable.ox.ac.uk> writes:
>
> >I wouldn't go so far as to say that
> >centrifugal force doesn't exist, since we can only observe forces from
> >their effects, and in a rotating frame of reference a centrifugal force
> >is most certainly observed
>
> Errrmm. No it`s not. Think of when you are in a car going round a bend. At
> first you appear move outwards from the bend, you are in actuall fact
Thanks for the explanation, but I knew that already. I really can't be
bothered to continue arguing fine points of physics, so I won't, but I stick by
my previous statement. My point was that although 'everybody knows' that
centrifugal force does not exist, it's a useful concept sometimes in rotating
frames of reference. Anyway, how do you define inertia? Relative to what?
Ben
--
Email: wadh...@sable.ox.ac.uk WWW: http://users.ox.ac.uk/~wadh0298
"You know my methods, Watson."
- 'The Memoirs of Sherlock Holmes. The Crooked Man',
Sir Arthur Conan Doyle
Erik Bjønnes
- Could it be because the walls is pointing to the middle of B5, and not to the outside?
- So that it would look like the floor is situated lying on the outisde walls of B5?
- Just wondering...
- Erik Bjønnes
Jake S Greenland
In article: <19970130190...@ladder01.news.aol.com>
mag...@aol.com writes:
> SO to reiterate. Centrifugal force exists neither from the standpoint
> of an outside viewer nor from someone inside the system.
>
Exactly, Centrifugal force in no way exists. It is, however, often
convenient to assume it does because it can make some calculations a
damn sight easier if you do.
Wibble on
Jake
--
--------------------------------------------------------
Ja...@Mythology.demon.co.uk Sojan@Discworld
"I am not a Frog, I am a Free Womble"
--------------------------------------------------------
Ian Jefferies
arl...@talia.girton.cam.ac.uk (Andy Lupini) wrote:
>In article <85464806...@eudoxus.demon.co.uk>,
> Ian.Je...@eudoxus.demon.co.uk (Ian Jefferies) writes:
>> arl...@talia.girton.cam.ac.uk (Andy Lupini) wrote:
>> Angular momentum is defined as the velocity multiplied by the closest distance
>> that the straight line path passes to the centre of rotation. The item
>> doesn't have to be attached to be a part of the calculation.
>No, for 2 reasons, can you guess what they are?
Which part are you saying no to: the definition of angular momentum or the
the attachment of items? The comment about "not being attached" in doing
conservation of mementum calculations is I believe valid: if the item is
unchanged in the before and after part of the conservation calculation then it
cancels out.
The last time I used angular momentum seriously was over 9 years ago. Thats
my excuse and I'm sticking to it...
Actually, now my brain cells are starting to kick in ... angular momentum is a
cross product (I'll pass on all the jokes) so it won't be a velocity vector
times a scalar, rather the cross product of the postion relative to the centre
of rotation, and velocity.
Erik Bjønnes
- Mike Hubbard wrote:
>
> On Tue, 28 Jan 97 20:48:39 GMT, linc...@sable.ox.ac.uk.anti-spam
-
> (rupert smith) wrote:
>
> >In article <1997012819...@dialup-11-28.netcomuk.co.uk>, agle...@netcomuk.co.uk (Alex Glennie) wrote:
>
-
> (Underwater garden sprinkler set to suck snipped)
>
-
> >lies! lies! lies!
> >it doesn't move at all.
> >
-
> >think angular momentum: the water going up the central pipe doesn't
have
> >much, so the rotor won't.
> >
> >>P.S. I guess it's possible you might need a huge super-duper computer to
> >>actually _prove_ the right answer
> >
> >alternatively, a basin of water and a bent straw. experimental physics
> >rules!
> >
-
> That's cheating! How are the universities going to get funding for
their
> supercomputers now? You can't make the students go back to the PC
> version of DOOM after they have played it on a Cray.
>
> If blowing water out of the sprinkler makes it go clockwise, then
> sucking it in will make it go anti-clockwise, won't it? If the rotors
> are at 12 O'clock and 6 O'clock, pointing towards 11 and 5, then that is
> where the water will be blown to or sucked from.
>
> Just like sucking or blowing on a toy windmill.
>
> Or is that what I am supposed to think?
- Or could it be ( now I'm not a physics student, but I'll take a chance...) that ist has some kind of a lock that prevents it going backwards? Or even perhaps a some sort of engine that uses the waters pressure to turn it around, and that it would just stop when it's rewersed?
- Erik
Andy Lupini
In article <85482282...@eudoxus.demon.co.uk>,
Ian.Je...@eudoxus.demon.co.uk (Ian Jefferies) writes:
>>> Angular momentum is defined as the velocity multiplied by the closest distance
>>> that the straight line path passes to the centre of rotation. The item
>>> doesn't have to be attached to be a part of the calculation.
>
>>No, for 2 reasons, can you guess what they are?
>
> Which part are you saying no to: the definition of angular momentum or the
> the attachment of items? The comment about "not being attached" in doing
> conservation of mementum calculations is I believe valid: if the item is
> unchanged in the before and after part of the conservation calculation then it
> cancels out.
You're probably right about that part, it was the 'ang. mom. is
defined as....' part that I meant was wrong.
> Actually, now my brain cells are starting to kick in ... angular momentum is a
> cross product (I'll pass on all the jokes) so it won't be a velocity vector
> times a scalar, rather the cross product of the postion relative to the centre
> of rotation, and velocity.
Yep, that's more like it, but you might also want to include a mass
somewhere as it is a momentum, after all. :)
*Disengages pedant mode*
Standard maths lecturer joke:
Q: What do you get if you cross a mosquito with a mountaineer?
A: Nothing, you can't cross a vector with a scalar.
Andy
rupert smith
In article <5d1pve$4...@lyra.csx.cam.ac.uk>, arl...@talia.girton.cam.ac.uk (Andy Lupini)
<E4F21...@liverpool.ac.uk> <5c5h1o$uk$1...@taliesin.netcom.net.uk>
<E4FAn...@liverpool.ac.uk> <32E6E8...@mad.scientist.com>
<5c6a7j$o5e$1...@taliesin.netcom.net.uk> <32E81C...@mad.scientist.com>
<5c8c1s$qdc$1...@taliesin.netcom.net.uk> <32E838...@mad.scientist.com>
<5c9fcd$r...@news.ox.ac.uk> <i7XbXTAD...@rauko.demon.co.uk>
<5cfn03$9b2$1...@taliesin.netcom.net.uk>
<OzbQVAA1...@rauko.demon.co.uk> <5ckj6o$e...@lyra.csx.cam.ac.uk>
<85464806...@eudoxus.demon.co.uk> <5cqr6d$a...@lyra.csx.cam.ac.uk>
<85482282...@eudoxus.demon.co.uk> wrote:
>> cross product (I'll pass on all the jokes) so it won't be a velocity vector
>> times a scalar, rather the cross product of the postion relative to the
> centre
>> of rotation, and velocity.
>
>Yep, that's more like it, but you might also want to include a mass
>somewhere as it is a momentum, after all. :)
strictly speaking L=P(cross)R
(or it might be R(cross)P, i'm still sleepy..)
anyway, the important thing is that it commutes with both.
rupert smith
In article <5d24ng$9...@news.ox.ac.uk>, linc...@sable.ox.ac.uk.anti-spam (rupert smith) wrote:
>In article <5d1pve$4...@lyra.csx.cam.ac.uk>, arl...@talia.girton.cam.ac.uk
> (Andy Lupini)
> <E4F21...@liverpool.ac.uk> <5c5h1o$uk$1...@taliesin.netcom.net.uk>
<snip>
> <85482282...@eudoxus.demon.co.uk> wrote:
sorry about all that, i'm not sure where it came from..
Ian Jefferies
linc...@sable.ox.ac.uk.anti-spam (rupert smith) wrote:
>In article <5d1pve$4...@lyra.csx.cam.ac.uk>, arl...@talia.girton.cam.ac.uk (Andy Lupini)
> wrote:
>>> cross product (I'll pass on all the jokes) so it won't be a velocity vector
>>> times a scalar, rather the cross product of the postion relative to the
>> centre
>>> of rotation, and velocity.
>>
>>Yep, that's more like it, but you might also want to include a mass
>>somewhere as it is a momentum, after all. :)
>strictly speaking L=P(cross)R
>(or it might be R(cross)P, i'm still sleepy..)
Definately R(cross)P, and you were right of course about the use of linear
momentum rather than velocity Andy. Thanks for pointing out the error.
It's amazing what you forget in ten years...
Alex Glennie
Martin Hardgrave <Mar...@deira.demon.co.uk> wrote:
> Probably. I'd hazard a guess that it:
> a continues to rotate the same way, as the water in the nozzle has
> some momentum which is transferred to the end of the head before going
> up the main tube, or
> b rotates the opposite way due to the moving liquid in front of
> the nozzle having a lower pressure which tends to "suck" the nozzle
> round, or
> c doesn't rotate at all.
One of the above is, indeed, ccccorrect.
Ian Jefferies
linc...@sable.ox.ac.uk.anti-spam (rupert smith) wrote:
>In article <5d24ng$9...@news.ox.ac.uk>, linc...@sable.ox.ac.uk.anti-spam (rupert smith) wrote:
>>In article <5d1pve$4...@lyra.csx.cam.ac.uk>, arl...@talia.girton.cam.ac.uk
>> (Andy Lupini)
>> <E4F21...@liverpool.ac.uk> <5c5h1o$uk$1...@taliesin.netcom.net.uk>
><snip>
>> <85482282...@eudoxus.demon.co.uk> wrote:
>sorry about all that, i'm not sure where it came from..
Its the message history: accurately placed the message in a thread even if
earlier messages have not yet been delivered to your news server. It
definately shouldn't be there.