Short Mars travel times at high speed.
Robert Clark
approach. If we have a 30 km/sec initial velocity to Mars, which might
be achievable with airbreathing(scramjet) or nuclear propulsion then
the travel time might be 23 days if you make a simplifying assumption
of a straight-line trip. However, the time required to make the
journey might be made significantly better than this 23 days.
The key fact is that the Earth itself has a 30 km/s velocity around
the Sun that can be used to give us an extra velocity boost toward the
orbit of Mars. In this new estimate I'll simplify the analysis by
assuming that at this high velocity and at the short travel time
achieved, the path will be essentially straight, rather than the
actual ellipse.
The famous Hohmann transfer orbit gives a minimal delta-v and energy
solution for traveling from one orbit to another but this is known for
its long travel times, 6 to 7 months for a Earth to Mars trip. We want
to shorten that for a manned trip to reduce the exposure to radiation
and to reduce the effects of long periods in zero-g.
I'll take the Earth orbit radius to be 150 million km and the Mars
orbit radius to be a little more than its distance at perihelion 210
million km. If we went in a tangential direction to Earth's orbit we
would have a total velocity toward the orbit of Mars of 60 km/s. The
problem here is that we would also have a longer straight-line travel
distance. This would result in the travel time being longer than
moving radially at 30 km/s. So the idea is to move at an optimal angle
that can use the Earth's orbital velocity while at the same time not
making the travel distance too long. See the image here for the
diagrams to illustrate the addition of velocities at an angle θ
(theta) and the travel distance at the angle θ calculations:
http://www.advancedphysics.org/forum/attachment.php?attachmentid=282
(may need to do a free registration at www.advancedphysics.org to
access the image.)
In the diagrams v is the total velocity, r is Earth's orbital radius,
R is Mars orbital radius and d is the straight-line travel distance.
Applying the law of cosines for the velocities gives for the total
velocity:
v^2 = 30^2 + 30^2 -2(30)(30)cos(180-θ) = 2(30^2)(1 + cos(θ))
So v = 30*sqrt(2(1 + cos(θ))
Applying the law of cosines for the travel distance gives the
equation:
R^2 = r^2 + d^2 -2(r)(d)cos(90 + θ) = r^2 + d^2 + 2(r)(d)sin(θ)
Solving for the travel distance d using the quadratic formula gives:
d = -rsin(θ) + sqrt(R^2 -(rcos(θ))^2)
I created a table using various angles θ in fractions of π (pi)
radians to find the shortest trip time:
θ (radians)| time (days)
-----------------------------
0 | 28.36
Ï€/2 | 16.4
Ï€/3 | 11.4
Ï€/4 | 12.9
Ï€/5 | 14.8
Ï€/6 | 16.4
Ï€/7 | 17.7
Ï€/8 | 18.7
Ï€/9 | 19.6
We see the shortest time at π/3 or 60 degrees is a surprising 11.4
days.
Quite a significant advantage than taking a 6 month long Hohmann
transfer orbit.
I'll assume like Robert Zubrin, author of the "Mars Direct" plan,
that aerobraking can be used to stop at Mars on arrival even at such
high speeds.
How reasonable is the assumption that at such high speeds and the
resulting short travel times the straight-line approximation is
accurate?
Bob Clark
Androcles
"Robert Clark" <rgrego...@yahoo.com> wrote in message
news:8cf7edc9-74e4-40a2...@m45g2000hsb.googlegroups.com...
(theta) and the travel distance at the angle è calculations:
http://www.advancedphysics.org/forum/attachment.php?attachmentid=282
(may need to do a free registration at www.advancedphysics.org to
access the image.)
In the diagrams v is the total velocity, r is Earth's orbital radius,
R is Mars orbital radius and d is the straight-line travel distance.
Applying the law of cosines for the velocities gives for the total
velocity:
v^2 = 30^2 + 30^2 -2(30)(30)cos(180-è) = 2(30^2)(1 + cos(è))
So v = 30*sqrt(2(1 + cos(è))
Applying the law of cosines for the travel distance gives the
equation:
R^2 = r^2 + d^2 -2(r)(d)cos(90 + è) = r^2 + d^2 + 2(r)(d)sin(è)
Solving for the travel distance d using the quadratic formula gives:
d = -rsin(è) + sqrt(R^2 -(rcos(è))^2)
I created a table using various angles è in fractions of ð (pi)
radians to find the shortest trip time:
è (radians)| time (days)
-----------------------------
0 | 28.36
ð/2 | 16.4
ð/3 | 11.4
ð/4 | 12.9
ð/5 | 14.8
ð/6 | 16.4
ð/7 | 17.7
ð/8 | 18.7
ð/9 | 19.6
We see the shortest time at ð/3 or 60 degrees is a surprising 11.4
days.
Quite a significant advantage than taking a 6 month long Hohmann
transfer orbit.
I'll assume like Robert Zubrin, author of the "Mars Direct" plan,
that aerobraking can be used to stop at Mars on arrival even at such
high speeds.
How reasonable is the assumption that at such high speeds and the
resulting short travel times the straight-line approximation is
accurate?
Bob Clark
==============================================
Incredibly naive...
Look at a shuttle and the size of the fuel tank needed to lift that
to Earth orbit so that it travels at 27,755 km per HOUR to match
speed with the ISS.
http://spaceflight.nasa.gov/realdata/tracking/
Now you want to travel at more than 3,600 times that, 27,755 km
per SECOND.
That means you have to lift into Earth orbit a rocket so huge it just
doesn't bear thinking about and THEN fire THAT up to travel to Mars.
But that's only half the story, you have to carry as much fuel mass at
the other end to stop it again.
Aero braking?
Columbia broke up using aero braking, and your ship is going
3,600 times faster than that.
And for the return trip...
What you need is a sleigh as fast as Santa's that can deliver all
the prezzies to all the kiddies all in one night. In other words, magic.
Do your sums again.
Spaceman
> (theta) and the travel distance at the angle è calculations:
>
> http://www.advancedphysics.org/forum/attachment.php?attachmentid=282
> (may need to do a free registration at www.advancedphysics.org to
> access the image.)
>
> In the diagrams v is the total velocity, r is Earth's orbital radius,
> R is Mars orbital radius and d is the straight-line travel distance.
> Applying the law of cosines for the velocities gives for the total
> velocity:
>
> Applying the law of cosines for the travel distance gives the
> equation:
>
>
> Solving for the travel distance d using the quadratic formula gives:
>
>
> radians to find the shortest trip time:
>
> -----------------------------
> 0 | 28.36
> ð/2 | 16.4
> ð/3 | 11.4
> ð/4 | 12.9
> ð/5 | 14.8
> ð/6 | 16.4
> ð/7 | 17.7
> ð/8 | 18.7
>
>
> We see the shortest time at ð/3 or 60 degrees is a surprising 11.4
> days.
> Quite a significant advantage than taking a 6 month long Hohmann
> transfer orbit.
> I'll assume like Robert Zubrin, author of the "Mars Direct" plan,
> that aerobraking can be used to stop at Mars on arrival even at such
> high speeds.
> How reasonable is the assumption that at such high speeds and the
> resulting short travel times the straight-line approximation is
> accurate?
Straight lines are always faster trips at the same speeds,
as long as you do actually go in the straight line.
Why not reduce the speed in half and double the time and we
still get there in about 23 days and don't have to worry about as
much aerobraking.
:)
--
James M Driscoll Jr
Spaceman
Robert Clark
...
> Incredibly naive...
> Look at a shuttle and the size of the fuel tank needed to lift that
> to Earth orbit so that it travels at 27,755 km per HOUR to match
> speed with the ISS.
> http://spaceflight.nasa.gov/realdata/tracking/
> Now you want to travel at more than 3,600 times that, 27,755 km
> per SECOND.
>
> That means you have to lift into Earth orbit a rocket so huge it just
> doesn't bear thinking about and THEN fire THAT up to travel to Mars.
> But that's only half the story, you have to carry as much fuel mass at
> the other end to stop it again.
> Aero braking?
> Columbia broke up using aero braking, and your ship is going
> 3,600 times faster than that.
> And for the return trip...
>
>
You misread the speed. Orbital velocity for low Earth orbit which the
shuttle reaches is about 7500 m/s or 7.5 km/sec. I'm suggesting a
speed of 4 times that 30 km/sec.
Every shuttle on return uses aerobraking under its normal aerodynamic
configuration to land. It was because a wing suffered severe damage
that caused Columbia to lose its normal configuration for landing that
caused it to break up.
Bob Clark
Sam Wormley
> The distance from Earth to Mars is about 60,000,000 km at closest
> approach. If we have a 30 km/sec initial velocity to Mars, which might
> be achievable with airbreathing(scramjet) or nuclear propulsion then
> the travel time might be 23 days if you make a simplifying assumption
> of a straight-line trip.
How are you going to to slow that craft down at Mars, short of vaporization
when plowing into the surface?
What is the cost of fuel/unit of mass of the craft comparing traditional
transfer orbits and your 23 "strait shot"?
Robert Clark
wrote:
You need the high speed to reduce the effect on the *overall* shape
of the trajectory by Earth's gravity, so the straight-line
approximation is accurate. At slower speeds the actual curved
elliptical path becomes dominant and you have to consider the effect
of that curved path on the travel time.
Think of throwing a ball horizontally. At a slow speed the curved
path is obvious. At a high speed the path becomes straighter and it is
able to reach a longer distance.
My guess is that at 15 km/sec this is so close to the escape velocity
of 11 km/sec that the curved elliptical path would become dominant and
the trip time would be only a little better than the curved Hohmann
transfer orbit time of 6 to 7 months.
Bob Clark
Sam Wormley
> My guess is that at 15 km/sec this is so close to the escape velocity
> of 11 km/sec that the curved elliptical path would become dominant and
> the trip time would be only a little better than the curved Hohmann
> transfer orbit time of 6 to 7 months.
>
>
> Bob Clark
Guess? Do the calculations!
Spaceman
Then it comes down to plotting a straight line once as far out of
the problem of escape velocity curving needed.
We should not be shooting just a ball, we should be shooting
a ball with controllable rockets on it.
:)
Once we can, we change course to "as close to straight line as possible".
:)
It all does come down to timing also.
There is no doubt that straight lines at same speeds will shorten trips.
So, finding the straight line when we can is the best method and
should shorten the trip simply.
Just as the shortest distance is a straight line.
and the shortest distance would also be a "Straight" path.
:)
60,000,000 kilometers.in a straight line at 15 km/sec = 4000000 seconds
about 1112 hrs
46 days if you actually travel a straight line at that speed,
and curves in the path of course will lengthen that time.
and increase in speed on the straight line shortens the trip.
Why do we take the long way when we really don't have to
if the "shooting" is timed correctly from the correct shooting
platform.
Willie...@gmail.com
To get an idea of what's going on you need an elementary knowledge of
the rocket equation - how rockets build speed - and an elementary
knowledge of orbital mechanics. The speed and transit times for
minimum energy orbits are calculated - and once that is understood, we
can then proceed to see what the benefits and costs of adding speed
are;
ROCKET EQUATION
The velocity of a rocket propelled projectile is given by the
Tsiolkovsky Equation;
Vf = Ve*LN(1/(1-u))
Where
Vf = final velocity
Ve= exhaust velocity
LN( ..) = natural logarithm function
u = propellant fraction.
So, if a rocket is 50% by weight propellant and its exhaust speed is 4
km/sec we can compute a final velocity for the rocket of;
Vf = 4.0 x LN(1/(1-0.5)) = 4.0 x LN(2) = 4.0 x 0.693147 = 2.772 km/
sec
If a rocket is 80% by weight propellant and its exhaust speed is 4.5
km/sec we can compute a final velocity for the rocket of;
Vf = 4.5 x LN(1/(1-0.8)) = 4.5 x LN(5) = 4.5 x 1.609438 = 7.242 km/
sec
Temperatures and pressures achievable with modern materials limit the
exhaust speeds of rocket engines. Strength to weight of materials
limit the amount of propellant a tank can carry. These are the design
parameters we have to work within.
There are several types of rocket engines that have been developed
over the years, and many practical systems proposed that are capable
of both high thrust and high performance.
1) solid propellant rockets - Ve = 2.5 km/sec
2) hypergolic liquid propellant rockets - Ve=3.2 km/sec
4) cryogenic liquid propellant rockets - Ve=4.5 km/sec
5) solid core nuclear thermal rocket with cryogenic propellant - Ve
= 9.0 km/sec
6) gas core nuclear thermal rocket with cryogenic propellant - Ve =
15.0 km/sec
7) nuclear pulse rockets- Ve = 25.0 km/sec
In recent years, after the advent of SDI, some have proposed replacing
the nuclear heat source in the last 3 rocket types with laser energy
beamed efficiently to the rocket - providing an increase of thrust to
weight.
ORBITAL MECHANICS
Orbital velocity on Earth surface is 6.5 km/sec to 8.2 km/sec.
Minimum Moon trajectory 10.9 km/sec (4 days)
Escape velocity on Earth surface is 11.2 km/sec
Minimum Mars Trajectory 11.8 km/sec (9 months)
This does not count air drag losses or gravity drag losses during
ascent. These add 1.2 km/sec to 2.0 km/sec depending. The ideal
final velocity for the Space Shuttle is 9.2 km/sec. So, any vehicle
with that delta vee capacity -in other words that Vf- can fly the same
flight envelope from Cape Canaveral Florida.
We can re-arrange the rocket equation to solve for propellant fraction
needed to attain these velocities
u = 1 - 1/EXP(Vf/Ve)
Orbital velocity Vf = 6.5 km sec
Ve = 2.5 km/sec ---> u = 0.9257
= 3.2 u = 0.8688
= 4.5 u = 0.7641
= 9.0 u = 0.5143
=15.0 u = 0.3517
=25.0 u = 0.2290
This shows that when the desired speed of a rocket exceeds the exhaust
velocity, its best to achieve that velocity in stages to reduce
overall mass.
A gas core or nuclear pulse rocket are practical ways to achieve very
high velocities - from the surface of the Earth. Upper stages need
only achieve 0.6 km/sec or more - to attain interplanetary speeds if
already on an escape trajectory put there by an existing rocket.
Since no existing rocket is large enough to send a manned payload to
escape velocities, when considering manned travel to Mars, we are
talking about multiple launches of existing rockets, and assembly on
orbit, OR, the construction of brand new rocket systems much larger
than the ones we use at present.
Both paths are extremely expensive - not as expensive as our invasion
of Afghanistan - but expensive nevertheless.
Interplanetary flight occurs along minimum energy trajectories -
called hohmann transfer orbits.
http://en.wikipedia.org/wiki/Hohmann_transfer
Which gives you minimum energy transfer delta vee - 'budgets'
http://en.wikipedia.org/wiki/Delta-v_budget#Interplanetary_budget
Basically add - 0.6 km/sec to the escape velocity from Earth's
surface, and you can get to mars.
You asked about transfer times. For that you need an introduction to
orbital mechanics.
http://www.braeunig.us/space/orbmech.htm
Which gets you to Androcles post - though I didn't check his math...
so, I can't warrant that.
However, Keplers third law of motion can be helpful here to understand
transfer times.
The squares of the orbital periods are equal to the cubes of the semi-
major axes.
So, the semi-major axis (radius) of Earth's orbit is 1 au, and its
period it 1 year.
1 x 1 = 1 x 1 x 1
The semi-major axis of Mars' orbit is 1.523679 AU and its period is
1.8808 years
1.523679 x 1.523679 x 1.523679 = 1.8808 x 1.8808
So, a hohmann transfer orbit has a perihelion of 1.0 AU and an
apohelion of 1.523679 AU add those up to get a major axis of 2.523679
- divide by 2 to get 1.261845 - cube that - (multiply by itself 3
times) to get 2.00091 and take the square root - to get 1.417454 years
- this is the time it takes to undergo a complete circuit - divide by
two to get the transfer time - 0.70872 years - multiply by 12 to get
8.504 months.
To understand all those sines and cosines understand what we're doing
- we're taking segments of the orbit, and calculating the transit time
over those segments - once we know the orbit
http://en.wikipedia.org/wiki/Image:Kepler_laws_diagram.svg
Generally speaking, if you accelerate a vehicle continuously along a
trajectory at accelerations that are large compared to the sun's local
gravity - at Earth and at Mars - then you can use straight line
approximations. If you are kicking the payload with a rocket blast at
the beginning of a journey - and the speed change due to the
accumulation of solar gravity influence is small during transit - then
you can use straight line approximations again.
The acceleration at the surface of the sun is 28.02 gees (274 m/s/s)
and the solar radius is 0.00452 AU (679,000 km) at 1 AU solar gravity
is reduced by a factor of 1/r^2 - or 1/48,800 th the gravity at the
surface. That's 574 micro gees (5.636 mm/s/s). At 1.52 AU that's
reduced to 248 micro gees.(2.435 mm/s/s)
Velocity is equal to acceleration times time. So, for our 8.5 months
= 22.35 million seconds
5 mm/s/s x 22.35 mega seconds = 126 km/sec
Over the course of the transit along a minimum energy orbit - you have
100 km/sec delta vee due to solar influence. A delta vee of
substantially higher velocity will look like a straight line and lose
little of its velocity.
Also, a transit at 1 gee - to maintain gee forces aboard the
spacecraft - will also look like a straight line - since solar
influence will be nil at 1/2 milligee.
So, here's an interesting calculation;
60 million km = 60 billion meters
1/2 this value is 30 billion meters
D = Vf^2 / 2*9.82
rearranging terms
Vf = SQRT(2* 30e9 * 9.82) = 767.59 km/sec
This takes a sort of ship that we haven't seriously considered yet -
one that uses say anti-matter to efficiently detonate fusion reactions
with exhaust velocities at 1,000 km/sec or more.
Where
Th = time of the transfer
Robert Clark
It's a two body problem in *two* dimensions, not one. Doable, but not
trivial.
The closest I've seen to it on the net is this presentation:
Basic of Space Flight: Orbital Mechanics:Orbit Altitude Changes.
http://www.braeunig.us/space/orbmech.htm#maneuver
But this takes the angle of departure as only tangential to the
initial orbit so you don't find the optimal angle to minimize the trip
time.
Bob Clark
Greg Neill
news:laadnUkgbIAAAvLV...@comcast.com
> Why do we take the long way when we really don't have to
> if the "shooting" is timed correctly from the correct shooting
> platform.
With enough fuel almost any desired path and transit
time (within reason) is possible. Unfortunately,
the energy cost is just too high to make it practical.
There is a time versus fuel tradeoff (or more technically
correct, a time versus energy tradeoff). The Hohmann
transfer orbit provides the least energy means of
getting from one orbit to another, barring tricks like
gravity assist manouvers where momentum is "stolen"
from other bodies.
Just getting fuel into space is expensive (where here
"expensive" means the cost in fuel mass). Look how
much fuel the shuttle burns just to end up in low
Earth orbit with practically empty tanks and having
discarded its solid rocket boosters.
Spaceman
> "Spaceman" <spac...@yourclockmalfunctioned.duh> wrote in message
> news:laadnUkgbIAAAvLV...@comcast.com
>
>> Why do we take the long way when we really don't have to
>> if the "shooting" is timed correctly from the correct shooting
>> platform.
>
> With enough fuel almost any desired path and transit
> time (within reason) is possible. Unfortunately,
> the energy cost is just too high to make it practical.
> There is a time versus fuel tradeoff (or more technically
> correct, a time versus energy tradeoff). The Hohmann
> transfer orbit provides the least energy means of
> getting from one orbit to another, barring tricks like
> gravity assist manouvers where momentum is "stolen"
> from other bodies.
So, all we really need to do is "trick" our ships
to steal Earths gravitational momentum to
get the speed we really want and time such
to be "as close to straight line" we also "really" want to shorten
the trip.
> Just getting fuel into space is expensive (where here
> "expensive" means the cost in fuel mass). Look how
> much fuel the shuttle burns just to end up in low
> Earth orbit with practically empty tanks and having
> discarded its solid rocket boosters.
And more silly is traveling around the globe to get
to the spot that is three feet away from you.
:)
Straight lines should use less total fuel in the final trip.
And.. tada... they always do.
:)
An energy trade off to not go straight is just stupid
in the end.
How does taking a longer trip, use less fuel in the end?
It would not, is the real answer.
:)
Sam Wormley
> On Jul 5, 12:06 pm, Sam Wormley <sworml...@mchsi.com> wrote:
>> Robert Clark wrote:
>>> My guess is that at 15 km/sec this is so close to the escape velocity
>>> of 11 km/sec that the curved elliptical path would become dominant and
>>> the trip time would be only a little better than the curved Hohmann
>>> transfer orbit time of 6 to 7 months.
>>> Bob Clark
>> Guess? Do the calculations!
>
> It's a two body problem in *two* dimensions, not one. Doable, but not
> trivial.
> The closest I've seen to it on the net is this presentation:
>
> Basic of Space Flight: Orbital Mechanics:Orbit Altitude Changes.
> http://www.braeunig.us/space/orbmech.htm#maneuver
Thanks for the online reference...
>
> But this takes the angle of departure as only tangential to the
> initial orbit so you don't find the optimal angle to minimize the trip
> time.
With proper timing, the optimal angle IS tangential, making the
most of the orbital velocity.
>
>
> Bob Clark
Androcles
| On Jul 5, 11:09 am, "Androcles" <Headmas...@Hogwarts.physics> wrote:
| ...
| > Incredibly naive...
| > Look at a shuttle and the size of the fuel tank needed to lift that
| > to Earth orbit so that it travels at 27,755 km per HOUR to match
| > speed with the ISS.
| > http://spaceflight.nasa.gov/realdata/tracking/
| > Now you want to travel at more than 3,600 times that, 27,755 km
| > per SECOND.
| >
| > That means you have to lift into Earth orbit a rocket so huge it just
| > doesn't bear thinking about and THEN fire THAT up to travel to Mars.
| > But that's only half the story, you have to carry as much fuel mass at
| > the other end to stop it again.
| > Aero braking?
| > Columbia broke up using aero braking, and your ship is going
| > 3,600 times faster than that.
| > And for the return trip...
| >
| >
|
| You misread the speed.
I gave the web page above. Here it is again.
http://spaceflight.nasa.gov/realdata/tracking/
Tell NASA they don't know what they are talking about, I'm sure they'd enjoy
the laugh.
| Orbital velocity for low Earth orbit which the
| shuttle reaches is about 7500 m/s or 7.5 km/sec. I'm suggesting a
| speed of 4 times that 30 km/sec.
Ok. 7710.05 mps over Japan as I look right now.
E = 1/2 mv^2
Four times the velocity, 16 times the energy.
You want to lift 16 FULL fuel tanks into orbit in preparation
for the Mars trip, outgoing, 16 more for stopping, and 32 for the return
trip.
Total 64 times as much. How much fuel will you need to lift the fuel?
Oh wait... you are going to develop a nuclear powered drive by 2020,
right?
NASA currently has Discovery, Atlantis, Endeavour. Challenger and
Columbia are no more. Let's build a bigger fleet, a hundred should
be enough. And the reason for this is to make a nice, comfortable
short trip to Mars for the crew so that the poor darlings don't get
bored on a long trip and have to ride an exercise bicycle to keep fit
as they do on the ISS.
| Every shuttle on return uses aerobraking under its normal aerodynamic
| configuration to land.
Yes.
| It was because a wing suffered severe damage
| that caused Columbia to lose its normal configuration for landing that
| caused it to break up.
|
| Bob Clark
Yes... and you plan on travelling 4 times faster and burn up 16 times
the kinetic energy in heating up the atmosphere after you've spent
64 times as much money as a slow boat to Mars will cost, all for
creature comfort.
How about if the oh-so-terrible government raises your taxes 64 times
to pay for this trip to a dust bowl in a desert? At least you'll be able to
plant a flag there, right?
Hey, I'm no greenie, I'm not a tree-hugger and I'm all for exploring
the planets, but hell man... have SOME common sense.
Uncle Al
>
> The distance from Earth to Mars is about 60,000,000 km at closest
> approach. If we have a 30 km/sec initial velocity to Mars, which might
> be achievable with airbreathing(scramjet) or nuclear propulsion then
> the travel time might be 23 days if you make a simplifying assumption
> of a straight-line trip.
You would be more credible if you knew anything about anything and
could do arithmetic. Shut up and calculate your momentum (mv) and
energy (mv^2/2) budgets for starters.
http://www.neofuel.com/optimum/
Gee, Rocky - if we had a gigawatt nuclear reactor with no radiation
shielding and an infinite supply of massless high momentum reaction
mass... A big LED thruster? Don't stand behind it!
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/lajos.htm#a2
Robert Clark
...
> > The closest I've seen to it on the net is this presentation:
>
> > Basic of Space Flight: Orbital Mechanics:Orbit Altitude Changes.
> >http://www.braeunig.us/space/orbmech.htm#maneuver
>
> Thanks for the online reference...
>
>
>
> > But this takes the angle of departure as only tangential to the
> > initial orbit so you don't find the optimal angle to minimize the trip
> > time.
>
> With proper timing, the optimal angle IS tangential, making the
> most of the orbital velocity.
>
>
>
> > Bob Clark
No. Just draw two concentric circles to see why it isn't. The
shortest distance from one to the other is radial, but then you a get
a smaller velocity boost from the Earths speed.
If you travel tangentially you get the biggest boost from the Earth's
speed but then the distance is longer.
For shortening the *time* an intermediate angle between these two
extremes is optimal, assuming the initial speed is high enough that
the straight-line approximation is accurate.
Bob Clark
Robert Clark
> Robert Clark wrote:
>
> > The distance from Earth to Mars is about 60,000,000 km at closest
> > approach. If we have a 30 km/sec initial velocity to Mars, which might
> > be achievable with airbreathing(scramjet) or nuclear propulsion then
> > the travel time might be 23 days if you make a simplifying assumption
> > of a straight-line trip.
>
> [snip crap]
>
> You would be more credible if you knew anything about anything and
> could do arithmetic. Shut up and calculate your momentum (mv) and
> energy (mv^2/2) budgets for starters.
>
> http://www.neofuel.com/optimum/
>
> Gee, Rocky - if we had a gigawatt nuclear reactor with no radiation
> shielding and an infinite supply of massless high momentum reaction
> mass... A big LED thruster? Don't stand behind it!
>
> --
> (Toxic URL! Unsafe for children and most mammals)http://www.mazepath.com/uncleal/lajos.htm#a2
You keep quoting that site without reading what it actually says.
This is for using low velocity exhaust without having to carry a huge
fuel load by having multiple large fuel dumps previously set up along
the way. The cost of having this much mass set up in multiple places
far out in space would already be prohibitive.
This would also be only a long travel time solution.
Bob Clark
Greg Neill
>> "Spaceman" <spac...@yourclockmalfunctioned.duh> wrote in message
>> news:laadnUkgbIAAAvLV...@comcast.com
>>
>>> Why do we take the long way when we really don't have to
>>> if the "shooting" is timed correctly from the correct shooting
>>> platform.
>>
>> With enough fuel almost any desired path and transit
>> time (within reason) is possible. Unfortunately,
>> the energy cost is just too high to make it practical.
>> There is a time versus fuel tradeoff (or more technically
>> correct, a time versus energy tradeoff). The Hohmann
>> transfer orbit provides the least energy means of
>> getting from one orbit to another, barring tricks like
>> gravity assist manouvers where momentum is "stolen"
>> from other bodies.
>
> So, all we really need to do is "trick" our ships
> to steal Earths gravitational momentum to
> get the speed we really want and time such
> to be "as close to straight line" we also "really" want to shorten
> the trip.
That would be nice if it were possible. But it's not
possible to steal enough momentum to approach a
straight line path in the solar system. The gravity
of the Sun makes sure that reasonable velocity
trajectories are all curved (conic sections).
Momentum is "stolen" by doing gravity assist trajectory
flybys. Problem is, each such fly by adds some velocity
to the craft, but you need to swing around and come back
for another pass to do it again. Each such pass uses
an unpowered orbit, so each successive pass would take
longer. Eventually the craft will reach Earth escape
velocity (less than 11 km/sec) and not come back at all.
>
>
>
>> Just getting fuel into space is expensive (where here
>> "expensive" means the cost in fuel mass). Look how
>> much fuel the shuttle burns just to end up in low
>> Earth orbit with practically empty tanks and having
>> discarded its solid rocket boosters.
>
> And more silly is traveling around the globe to get
> to the spot that is three feet away from you.
> :)
> Straight lines should use less total fuel in the final trip.
> And.. tada... they always do.
Not in space though, where unpowered motion is the
most fuel efficient way to go -- just travel
inertially, free-falling from one place to another
after an initial shove. These paths are all conic
sections.
> :)
> An energy trade off to not go straight is just stupid
> in the end.
Such is life in the real world. In space, near large
masses where things orbit, straight line paths are
amongst the most expensive ways to go!
> How does taking a longer trip, use less fuel in the end?
Because you don't burn any for most of the trip; you
just glide along on inertia.
> It would not, is the real answer.
Wrong.
Spaceman
Fighting such a small gravitational pull of the Sun,
on such a small object as a spaceship is not as hard
as one wants to make it.
:)
> Momentum is "stolen" by doing gravity assist trajectory
> flybys. Problem is, each such fly by adds some velocity
> to the craft, but you need to swing around and come back
> for another pass to do it again. Each such pass uses
> an unpowered orbit, so each successive pass would take
> longer. Eventually the craft will reach Earth escape
> velocity (less than 11 km/sec) and not come back at all.
And if such speed is achieved the straight path
becomes more possible to even fight the Suns pull
also.
Let me ask you this,
If a curved path is better,
Why do we try to basically shoot "striaght" upward first at all?
According to the curve sillyness, we should be shooting sideways
first.
:)
> Not in space though, where unpowered motion is the
> most fuel efficient way to go -- just travel
> inertially, free-falling from one place to another
> after an initial shove. These paths are all conic
> sections.
Enough speed will allow the straight path with an
unpowered flight that will be just enough to negate
the Sun's pull.
> Such is life in the real world. In space, near large
> masses where things orbit, straight line paths are
> amongst the most expensive ways to go!
No,
Try thinking of a trip to the top of a mountian.
Would you burn more fuel by taking a curved path to the top,
or a striaght line towards the top?
Do you think an airplane will burn more fuel making a curved
line to the top than it would by traveling a straight shot towards it?
> Because you don't burn any for most of the trip; you
> just glide along on inertia.
the inertia is the savings.
If the amount of inertia was used in a straight line.
It would be faster to get there and burn less fuel in the end.
Again
would you shoot a bullet to the side to get higher altitude
from the "inertia" or stright up?
>> It would not, is the real answer.
>
> Wrong.
No,
And cars an airplanes prove you are wrong
every single day.
oh ya.. and so don't rockets.
Greg Neill
It is what it is. The acceleration due to the Sun's
gravity depends on position (distance from the Sun),
not velocity.
>
>> Momentum is "stolen" by doing gravity assist trajectory
>> flybys. Problem is, each such fly by adds some velocity
>> to the craft, but you need to swing around and come back
>> for another pass to do it again. Each such pass uses
>> an unpowered orbit, so each successive pass would take
>> longer. Eventually the craft will reach Earth escape
>> velocity (less than 11 km/sec) and not come back at all.
>
> And if such speed is achieved the straight path
> becomes more possible to even fight the Suns pull
> also.
Again, the acceleration due to the Sun's gravity
depends on position, not velocity.
> Let me ask you this,
> If a curved path is better,
> Why do we try to basically shoot "striaght" upward first at all?
> According to the curve sillyness, we should be shooting sideways
> first.
Horizontal takeoff would be ideal if it weren't
for the atmosphere. The atmophere causes drag,
so you want to get through it as quickly as
possible. If you look at the actual paths of
rockets taking off, you'll see that they start
off going straight up, then bank over as they
get higher, eventually travelling parallel to
the Earth's surface directly below them as they
reach orbital height.
> :)
>
>> Not in space though, where unpowered motion is the
>> most fuel efficient way to go -- just travel
>> inertially, free-falling from one place to another
>> after an initial shove. These paths are all conic
>> sections.
>
> Enough speed will allow the straight path with an
> unpowered flight that will be just enough to negate
> the Sun's pull.
No. You can approach stright line, but never achieve
it. The fastest thing around is light, and even it's
path is made curved by the Sun's gravity.
>> Such is life in the real world. In space, near large
>> masses where things orbit, straight line paths are
>> amongst the most expensive ways to go!
>
> No,
> Try thinking of a trip to the top of a mountian.
> Would you burn more fuel by taking a curved path to the top,
> or a striaght line towards the top?
> Do you think an airplane will burn more fuel making a curved
> line to the top than it would by traveling a straight shot towards it?
The devil is in the details. Running a non 100%
efficient engine for longer time against the vagaries of
friction on the path will ultimately use more energy
for a longer path.
The ideal, least energy path to the top of the mountain
would be to be fired from a cannon. Expend the requisite
energy in one efficient burst to impart the kinetic energy
corresponding to the potential energy difference between
the bottom and top of the mountain.
In space there is (essentially) no friction. So unpowered
motion is free.
>
>
>> Because you don't burn any for most of the trip; you
>> just glide along on inertia.
>
> the inertia is the savings.
The fuel is the savings. Intertia is, well, inertia.
'
> If the amount of inertia was used in a straight line.
Inertia isn't 'used'. Inertia is the tendancy of a
body to resist change in momentum.
> It would be faster to get there and burn less fuel in the end.
Burning no fuel is always cheapest. Burning just the
needed fuel as efficiently as possible is the next
cheapest.
> Again
> would you shoot a bullet to the side to get higher altitude
> from the "inertia" or stright up?
Staight up, of course, if highest altitude was your goal.
Air friction determines the ideal initial angle. If there
were no air friction, and the bullet could pass freely
through the Earth (obviously a hypothetical situation) any
angle would do as the bullet would go into an orbit with a
major axis fixed by the initial position and velocity. It
would reach the same height above the "surface" eventually
(apogee).
Our notions of efficient paths on Earth are misleading
when it comes to space travel, since down here we tend
to take into account the "difficulty" of terrain and
air friction and other losses. In space the cheapest
way to travel is to just coast after making the most
efficient expenditure of fuel that we can (usually
meaning a short efficient rocket firing).
In space you don't have to worry about air friction.
Spaceman
> It is what it is. The acceleration due to the Sun's
> gravity depends on position (distance from the Sun),
> not velocity.
I never said its pull depends on the objects velocity,
I said the velocity if high enough negates that pull.
> Horizontal takeoff would be ideal if it weren't
> for the atmosphere. The atmophere causes drag,
> so you want to get through it as quickly as
> possible. If you look at the actual paths of
> rockets taking off, you'll see that they start
> off going straight up, then bank over as they
> get higher, eventually travelling parallel to
> the Earth's surface directly below them as they
> reach orbital height.
If you want to get there as quickly as possible,
You travel a straight line.
wind or not,
To get to a higher altitude, a straight line is the shorter
path always.
Ans a shorter path will burn less fuel in the end.
again.
Traveling to a mountain top proves it.
Why don't you think about such?
>> Enough speed will allow the straight path with an
>> unpowered flight that will be just enough to negate
>> the Sun's pull.
>
> No. You can approach stright line, but never achieve
> it. The fastest thing around is light, and even it's
> path is made curved by the Sun's gravity.
Because light doesn ot have built in rockets.
So..it can't fight gravity,
Spaceships can.
Again you seem to like to ignore that the shortest
distance is a striaght line and the least fuel burned
comes from a straight line also.
> The devil is in the details. Running a non 100%
> efficient engine for longer time against the vagaries of
> friction on the path will ultimately use more energy
> for a longer path.
Correct, longer path = more fuel use.
always.
Now think about that for once.
:)
> The ideal, least energy path to the top of the mountain
> would be to be fired from a cannon. Expend the requisite
> energy in one efficient burst to impart the kinetic energy
> corresponding to the potential energy difference between
> the bottom and top of the mountain.
The ideal path would be a straight line.
Again, you seem to ignore the fact that a straight line
uses the least fuel.
> In space there is (essentially) no friction. So unpowered
> motion is free.
And if enough fuel that brings you to that curve
is used in a straight path instead, you will be getting there
faster for the same amount of fuel.
:)
> The fuel is the savings. Intertia is, well, inertia.
Fuel is always saved by traveling a straight line.
Do you actually think taking a longer path
uses less fuel?
Again, I invite you to prove such to a trip the the
mountain top.
:)
> Burning no fuel is always cheapest. Burning just the
> needed fuel as efficiently as possible is the next
> cheapest.
Again, shorter paths will burn less fuel.
You seem to want to ignore that fact.
> Staight up, of course, if highest altitude was your goal.
> Air friction determines the ideal initial angle. If there
> were no air friction, and the bullet could pass freely
> through the Earth (obviously a hypothetical situation) any
> angle would do as the bullet would go into an orbit with a
> major axis fixed by the initial position and velocity. It
> would reach the same height above the "surface" eventually
> (apogee).
> Our notions of efficient paths on Earth are misleading
> when it comes to space travel, since down here we tend
> to take into account the "difficulty" of terrain and
> air friction and other losses. In space the cheapest
> way to travel is to just coast after making the most
> efficient expenditure of fuel that we can (usually
> meaning a short efficient rocket firing).
>
> In space you don't have to worry about air friction.
In reality the shortest path is the least fuel burned.
and not once is there 0 gravity towards Mars.
so the straight path will still win for least fuel burned
once speed is achieved.
:)
Again,
If you don't want to think about taking the trip to
the top of the mountain and the trip back will
be "basically free" then you don't get the fact about
shortest distance will use least fuel all the time.
:)
spudnik
(brachistochrone or tautachrone per Liebniz' calculus.
I always thought that the efficient "burn" of fuel,
required burningot throughout the trip,
halfway in acceleration & deceleration. it seems that
such speeds on the approach to Mars,
might strip its atmosphere, before you stopped (or,
the aerobrakeing'd take longer than the trip, or
cover a significant portion of Mars with the 'chute !-)
> If you don't want to think about taking the trip to
> the top of the mountain and the trip back will
> be "basically free" then you don't get the fact about
> shortest distance will use least fuel all the time.
thus:
I should add that these may be really important features
in the bugset:
your cones are not generally quadric surfaces, and
Joe's facets are not generally convex.
just because you guys tried to hide that, or
merely averted your eyes
in the patent-pending or paper-pretending,
doesn't mean it isn't "all good."
like, I'd call yours,"post-whacked dickonoids, " or
"flying funny shingles" or some thing; seriously ...
after a *lot* more of math, or at least systematic experiment
(that is to say, with a good write-up of observables,
not just a virtual tour of the latter-day shack).
>http://uspto.gov/web/patents/patog/week26/OG/html/1331-4/US07389612-20080624.html
--Seargent Barracks Soros McCheeny Pepper,
"Give jihad a chance in The Sudan, Rhodesia, and
other former colonial moments -- Yahoo!TM;
you're going to feel my computerized draft,
boys'n'girls: NO AMERICAN MIDDLESCHOOLER LEFT BEHIND;
NO RHODESIA SCHOLARS IN HARM'S WAY!"
http://larouchepub.com/lar/2008/3526lar_soros_pamph.html
http://larouchepub.com/other/2008/3526save_nations_parasites.html
http://larouchepub.com/other/2008/3526zim_brit_op.html
Spaceman
> the quickest path between two places is a slalom
> (brachistochrone or tautachrone per Liebniz' calculus.
The why do downhill skiers make it past the finish line first
and faster.
:)
The quickest path at a set speed is a straight line.
The least fuel burned at the same set speed is also a straight path.
Why are such simple facts ignored so well to come
up with this curved path crap?
Such ignorance of reality is amazing.
LOL
Greg Neill
>> It is what it is. The acceleration due to the Sun's
>> gravity depends on position (distance from the Sun),
>> not velocity.
>
> I never said its pull depends on the objects velocity,
> I said the velocity if high enough negates that pull.
Velocity cannot negate force. They are different
things. The force remains. What can happen is
that the relative amount of deflection for a given
length of path can be made smaller.
>
>
>> Horizontal takeoff would be ideal if it weren't
>> for the atmosphere. The atmophere causes drag,
>> so you want to get through it as quickly as
>> possible. If you look at the actual paths of
>> rockets taking off, you'll see that they start
>> off going straight up, then bank over as they
>> get higher, eventually travelling parallel to
>> the Earth's surface directly below them as they
>> reach orbital height.
>
> If you want to get there as quickly as possible,
> You travel a straight line.
Getting there is not the same thing as staying there.
If you went straight up and turned off your engine,
you'd fall straight back down. You need a velocity
parallel to the surface if you want to remain in
orbit.
> wind or not,
> To get to a higher altitude, a straight line is the shorter
> path always.
> Ans a shorter path will burn less fuel in the end.
> again.
> Traveling to a mountain top proves it.
> Why don't you think about such?
Been there, done that.
>>> Enough speed will allow the straight path with an
>>> unpowered flight that will be just enough to negate
>>> the Sun's pull.
>>
>> No. You can approach stright line, but never achieve
>> it. The fastest thing around is light, and even it's
>> path is made curved by the Sun's gravity.
>
> Because light doesn ot have built in rockets.
> So..it can't fight gravity,
> Spaceships can.
> Again you seem to like to ignore that the shortest
> distance is a striaght line and the least fuel burned
> comes from a straight line also.
Fighting gravity takes fuel. Coasting takes none.
>
>
>> The devil is in the details. Running a non 100%
>> efficient engine for longer time against the vagaries of
>> friction on the path will ultimately use more energy
>> for a longer path.
>
> Correct, longer path = more fuel use.
> always.
Not in space.
> Now think about that for once.
Been there, done that, got the T-shirt.
>
>> The ideal, least energy path to the top of the mountain
>> would be to be fired from a cannon. Expend the requisite
>> energy in one efficient burst to impart the kinetic energy
>> corresponding to the potential energy difference between
>> the bottom and top of the mountain.
>
> The ideal path would be a straight line.
> Again, you seem to ignore the fact that a straight line
> uses the least fuel.
It would take an infinite amount of energy to traverse
a straight line free-fall path through a gravitational
field. The minimum possible energy expenditure is
dictated by the potental energy difference in the
starting and ending locations. That can only be achieved
in a frictionless environment with a perfect engine.
The next best thing is a ballistic trajectory.
>> In space there is (essentially) no friction. So unpowered
>> motion is free.
>
> And if enough fuel that brings you to that curve
> is used in a straight path instead, you will be getting there
> faster for the same amount of fuel.
No, you must burn *more* fuel to follow a straight line
path if you're countering acceleration rather than just
letting the acceleration take you along.
>> The fuel is the savings. Intertia is, well, inertia.
>
> Fuel is always saved by traveling a straight line.
> Do you actually think taking a longer path
> uses less fuel?
Absolutely, in orbital manouvering. It is mathematically
proven as well as empirically observed. Remember,
with a ballistic trajectory you run the engine for only
a very brief time, whereas to constantly "drive" a
non inertial path you need to run the engine constantly.
> Again, I invite you to prove such to a trip the the
> mountain top.
Suppose you need to get to the other side of a mountain.
You can either coast around it on a perfectly flat
road, or drive straight up and straight down again.
What do you think will take more fuel?
>> Burning no fuel is always cheapest. Burning just the
>> needed fuel as efficiently as possible is the next
>> cheapest.
>
> Again, shorter paths will burn less fuel.
> You seem to want to ignore that fact.
Not if the path length is irrelevant to the fuel
consumption. Remember, for 99% of the trip you
need burn *no* fuel for a ballistic orbital path.
>
>> Staight up, of course, if highest altitude was your goal.
>> Air friction determines the ideal initial angle. If there
>> were no air friction, and the bullet could pass freely
>> through the Earth (obviously a hypothetical situation) any
>> angle would do as the bullet would go into an orbit with a
>> major axis fixed by the initial position and velocity. It
>> would reach the same height above the "surface" eventually
>> (apogee).
>
>
>> Our notions of efficient paths on Earth are misleading
>> when it comes to space travel, since down here we tend
>> to take into account the "difficulty" of terrain and
>> air friction and other losses. In space the cheapest
>> way to travel is to just coast after making the most
>> efficient expenditure of fuel that we can (usually
>> meaning a short efficient rocket firing).
>>
>> In space you don't have to worry about air friction.
>
> In reality the shortest path is the least fuel burned.
Not if you don't have to burn fuel over the whole path.
Then the path length is irrelevant.
> and not once is there 0 gravity towards Mars.
> so the straight path will still win for least fuel burned
> once speed is achieved.
No. You'll be contantly fighting gravity to maintain
a straight line, which is a waste of fuel for the sake
of geoemtry, whereas letting the ship coast along
without burning fuel costs nothing but time.
> Again,
> If you don't want to think about taking the trip to
> the top of the mountain and the trip back will
> be "basically free" then you don't get the fact about
> shortest distance will use least fuel all the time.
James, I've done the math and solved for the most
efficient, least fuel path. It goes by the name
of the Hohmann Transfer (there are special cases
where another trajectory consisting of a pair of
elliptical transfers can be slightly more efficient)
Look it up.
Greg Neill
You're ignoring the effect of gravity and asssuming
that the ship has a road to run on. Skiiers must
follow the mountain terrain and are not free to
choose their own 3D trajectory. If they were, and
could ignore the ski gates, they would follow a
curved path sa "spudnik" wrote. It's a classic
2nd year calculus problem.
Spaceman
> "Spaceman" <spac...@yourclockmalfunctioned.duh> wrote in message
> news:8tudnf3dj7bIVPLV...@comcast.com
>> Greg Neill wrote:
>>> It is what it is. The acceleration due to the Sun's
>>> gravity depends on position (distance from the Sun),
>>> not velocity.
>>
>> I never said its pull depends on the objects velocity,
>> I said the velocity if high enough negates that pull.
>
> Velocity cannot negate force.
So wrong, it is not even funny that you stated that.
Please prove a longer path will take less energy to travel than a
shorter path Greg,
When you do, tell the car companies that so they can laugh
also.
:)
<snipped rest of ignorance of a known fact about paths vs energy
use. >
Spaceman
> "Spaceman" <spac...@yourclockmalfunctioned.duh> wrote in message
> news:tIadnUsaw4h4S_LV...@comcast.com
>> spudnik wrote:
>>> the quickest path between two places is a slalom
>>> (brachistochrone or tautachrone per Liebniz' calculus.
>>
>> The why do downhill skiers make it past the finish line first
>> and faster.
>> :)
>> The quickest path at a set speed is a straight line.
>> The least fuel burned at the same set speed is also a straight path.
>> Why are such simple facts ignored so well to come
>> up with this curved path crap?
>> Such ignorance of reality is amazing.
>
> You're ignoring the effect of gravity and asssuming
> that the ship has a road to run on.
No I am not.
I am ignoring your ignorance of a longer path needed more
fuel than a shorter path would.
And sadly, you are ignoring that fact completely to come
up with your "curved path nonsense".
:)
Hint Greg:
There is no spot between here and Mars where there is no
gravity.
so even when you are "freefalling" you are taking a longer
path and will need to make it up with fuel again to
once again fight the gravity taking the longer trip.
> Skiiers must
> follow the mountain terrain and are not free to
> choose their own 3D trajectory. If they were, and
> could ignore the ski gates, they would follow a
> curved path sa "spudnik" wrote. It's a classic
> 2nd year calculus problem.
It's a joke and you don't seem to know how stupid it really
is in reality.
The shortest distance is the straight line, and the straght
line (even when fighting gravity) will burn less fuel.
Greg Neill
>> "Spaceman" <spac...@yourclockmalfunctioned.duh> wrote in message
>> news:8tudnf3dj7bIVPLV...@comcast.com
>>> Greg Neill wrote:
>>>> It is what it is. The acceleration due to the Sun's
>>>> gravity depends on position (distance from the Sun),
>>>> not velocity.
>>>
>>> I never said its pull depends on the objects velocity,
>>> I said the velocity if high enough negates that pull.
>>
>> Velocity cannot negate force.
>
> So wrong, it is not even funny that you stated that.
>
> Please prove a longer path will take less energy to travel than a
> shorter path Greg,
Get yourself almost any text on orbital mechanics.
> When you do, tell the car companies that so they can laugh
> also.
Cars are not spaceships. Cars are constrained to run on
the roads provided.
If roads were carved as curved tunnels from point to
point on the earth, with the right curve a car could
get from any point to any other point on the surface
using practically no fuel! Of course, every trip
would take a fixed time of 42 minutes, which might
prove inconvenient if you just wanted to go down
the block.
Spaceman
> "Spaceman" <spac...@yourclockmalfunctioned.duh> wrote in message
> news:68udnQw_J9EpQfLV...@comcast.com
>> Greg Neill wrote:
>>> "Spaceman" <spac...@yourclockmalfunctioned.duh> wrote in message
>>> news:8tudnf3dj7bIVPLV...@comcast.com
>>>> Greg Neill wrote:
>>>>> It is what it is. The acceleration due to the Sun's
>>>>> gravity depends on position (distance from the Sun),
>>>>> not velocity.
>>>>
>>>> I never said its pull depends on the objects velocity,
>>>> I said the velocity if high enough negates that pull.
>>>
>>> Velocity cannot negate force.
>>
>> So wrong, it is not even funny that you stated that.
>>
>> Please prove a longer path will take less energy to travel than a
>> shorter path Greg,
>
> Get yourself almost any text on orbital mechanics.
Why?
so I can read crap that is proven wrong about straight
paths vs curved paths and become brainwashed like
you to think a longer path could take less energy?
No thanks.
:)
LOL
> Cars are not spaceships. Cars are constrained to run on
> the roads provided.
Energy is energy and shorter paths are shorter paths.
Car or spaceship makes no difference in the end.
The shortest path will always use the least fuel.
> If roads were carved as curved tunnels from point to
> point on the earth, with the right curve a car could
> get from any point to any other point on the surface
> using practically no fuel! Of course, every trip
> would take a fixed time of 42 minutes, which might
> prove inconvenient if you just wanted to go down
> the block.
Wow,
you sure are lost with your curvature stuff.
A straight tunnel would be the least fuel needed
from point A to point B.
a curved tunnel up or down from such would burn more fuel
in the end no matter what you may think.
You simply can not grasp that simple fact huh?
Greg Neill
No. You give the ship the necessary impulse at the
beginning of the trip and allow it to coast all the
way to the vicinity of Mars' orbit. Then apply another
short burst to circularize the orbit to coincide with
that of Mars.
>
>> Skiiers must
>> follow the mountain terrain and are not free to
>> choose their own 3D trajectory. If they were, and
>> could ignore the ski gates, they would follow a
>> curved path sa "spudnik" wrote. It's a classic
>> 2nd year calculus problem.
>
> It's a joke and you don't seem to know how stupid it really
> is in reality.
Reality is not stupid. Ignoring it by trying to apply
your backyard experience is.
> The shortest distance is the straight line, and the straght
> line (even when fighting gravity) will burn less fuel.
No. The shortest distance may be a straight line, but
it's not always the most energy efficient. Remember,
you don't need to keep your engine running the whole
way.
Greg Neill
>> "Spaceman" <spac...@yourclockmalfunctioned.duh> wrote in message
>> news:68udnQw_J9EpQfLV...@comcast.com
>>> Greg Neill wrote:
>>>> "Spaceman" <spac...@yourclockmalfunctioned.duh> wrote in message
>>>> news:8tudnf3dj7bIVPLV...@comcast.com
>>>>> Greg Neill wrote:
>>>>>> It is what it is. The acceleration due to the Sun's
>>>>>> gravity depends on position (distance from the Sun),
>>>>>> not velocity.
>>>>>
>>>>> I never said its pull depends on the objects velocity,
>>>>> I said the velocity if high enough negates that pull.
>>>>
>>>> Velocity cannot negate force.
>>>
>>> So wrong, it is not even funny that you stated that.
>>>
>>> Please prove a longer path will take less energy to travel than a
>>> shorter path Greg,
>>
>> Get yourself almost any text on orbital mechanics.
>
> Why?
> so I can read crap that is proven wrong about straight
> paths vs curved paths and become brainwashed like
> you to think a longer path could take less energy?
> No thanks.
Ignorance is not a way of knowing things, as you
are very plainly demonstrating.
>> Cars are not spaceships. Cars are constrained to run on
>> the roads provided.
>
> Energy is energy and shorter paths are shorter paths.
> Car or spaceship makes no difference in the end.
> The shortest path will always use the least fuel.
>
>
>> If roads were carved as curved tunnels from point to
>> point on the earth, with the right curve a car could
>> get from any point to any other point on the surface
>> using practically no fuel! Of course, every trip
>> would take a fixed time of 42 minutes, which might
>> prove inconvenient if you just wanted to go down
>> the block.
>
> Wow,
> you sure are lost with your curvature stuff.
> A straight tunnel would be the least fuel needed
> from point A to point B.
No. With a tunnel that descends from the starting
point and ascends to the termination, the car could
coast the whole way with no fuel at all if there
were no friction. Like dropping a marble down the
side of a curved bowl, it would roll down one side
and back up the other.
> a curved tunnel up or down from such would burn more fuel
> in the end no matter what you may think.
> You simply can not grasp that simple fact huh?
You obviously know nothing about conservation of energy.
Spaceman
> No. The shortest distance may be a straight line, but
> it's not always the most energy efficient. Remember,
> you don't need to keep your engine running the whole
> way.
Let me correct a goof up I have said,
I used the "fuel" word where I should have used
the energy word.
The shortest distance being a straight line will
always take the least amount of energy,
and the longer the distance will take more energy
always.
and
A slingshot effect is simply grabbing that extra energy needed
for the longer trip.
Once you have used up the slingshot,
you are back to square one.
The shortest distance is best again, unless you
can get another slingshot.
but sadly curving slowly outward away from the sun,
by taking the "curved path to Mars" is not "gathering energy"
and hence the longer path doing such is stupid and will
not use less energy and in fact would need more energy
coming from the spaceship to do such.
So,
I admit my use of the word fuel was wrong,
but the energy will always be more used for the longer
path.
:)
Now show me a circular path away from gravity that
will use less energy than the straight path would Greg.
:)
Spaceman
> No. With a tunnel that descends from the starting
> point and ascends to the termination, the car could
> coast the whole way with no fuel at all if there
> were no friction. Like dropping a marble down the
> side of a curved bowl, it would roll down one side
> and back up the other.
I admit, I should not have used the word "fuel"
I should have only used the word energy.
so.
Please show me a longer path from A to B that does
not require more energy to travel than a shorter path
from A to B would.
:)
Greg Neill
A straight path from one side of the bowl to the
other, across the open mouth of the bowl (not
around the rim). You would have to burn fuel
continuously to hover while heading across. The
marble, rolling down the bowl and up again, can
get there without burning any fuel at all.
Greg Neill
In space travel, fuel is intimately associated with
energy. That's why I didn't take you to task for
the usage in the first place. It's common to equate
fuel with energy expenditures. Further, spacecraft
manouvers are usually calculated in terms of "delta v",
that is, the required velocity changes, which in turn
require a certain energy and certain amount of fuel for
a given craft. So, for example, departing Earth orbit
to begin a trip to Mars may require a delta V of 4km/sec.
At the Mars end, the ship may need to make a delta V of
3 km/sec to circularize into Mars' orbit.
>
> Now show me a circular path away from gravity that
> will use less energy than the straight path would Greg.
Any elliptical orbit section that gets you where you
want to go will take less energy than a straight path
would, simply because you don't need to be burning
energy contantly just to maintain a straight path (note
that much of the energy required to maintain a straight
path will *not* be getting you to your destination
quicker... it will be at an angle to your direction of
motion, in the direction opposing the gravitational
acceleration).
Spaceman
Did you just read what I stated or not..
You just proved you did not.
The longer trip uses more energy.
(I corrected myself about the use of the word fuel)
but apparently you did not even read the paragraphs
you just replied to.
:)
Your marble used more energy for the longer trip.
unless somehow you think it violates the conservation
of energy.
Spaceman
> In space travel, fuel is intimately associated with
> energy. That's why I didn't take you to task for
> the usage in the first place. It's common to equate
> fuel with energy expenditures. Further, spacecraft
> manouvers are usually calculated in terms of "delta v",
> that is, the required velocity changes, which in turn
> require a certain energy and certain amount of fuel for
> a given craft. So, for example, departing Earth orbit
> to begin a trip to Mars may require a delta V of 4km/sec.
> At the Mars end, the ship may need to make a delta V of
> 3 km/sec to circularize into Mars' orbit.
>
>>
>> Now show me a circular path away from gravity that
>> will use less energy than the straight path would Greg.
>
> Any elliptical orbit section that gets you where you
> want to go will take less energy than a straight path
> would, simply because you don't need to be burning
> energy contantly just to maintain a straight path (note
> that much of the energy required to maintain a straight
> path will *not* be getting you to your destination
> quicker... it will be at an angle to your direction of
> motion, in the direction opposing the gravitational
> acceleration).
You truly like to ignore the energy given by the
slingshot huh?
That is pretty sad ignoring all that energy you
used to take the longer path.
You did nto have to store it true, but you still
needed it.
Again,
once the slingshot is done.
The uphill trip to Mars is still there.
You better have enough energy stored now.
I know a straight path up the hill will need less
energy than the spiral path you wish to take.
:)
It is pretty sad you won't admit that fact at all.
Apparently you do have some form of violation
of the conservation of energy occuring in your
magical uphill spiral path.
:)
Greg Neill
No it doesn't. It makes an even trade of potential
energy for kinetic on the way down, then reverses
that and trades its kinetic energy for potential
energy on the way back up. Net change, zero. And
no engine required to do it either, Nature provides
the service for free. So no energy expenditure.
The other guy, though, has to burn a whole lot of
unrecoverable energy resources to take the shorter
trip.
> (I corrected myself about the use of the word fuel)
> but apparently you did not even read the paragraphs
> you just replied to.
For space travel it actually makes sense to talk in
terms of fuel requirements for manouvers. Fuel is
doled out in units of the Delta-V produced (so a
mission might have a total Delta-V budget of, say
15 km/sec corresponding to its entire fuel capacity).
> Your marble used more energy for the longer trip.
> unless somehow you think it violates the conservation
> of energy.
Nope, it obeyed conservation and got all its potential
energy back for a net change of zero, and no fuel
burned.
Greg Neill
What energy is that? In order to gain energy by
slingshot manouver around the Earth, you first need
to spend several months travelling away from the
Earth and picking up speed to lap the Earth in its
orbit (thus coming up on it from behind). If you're
lucky and you do the manouver just right, you can
almost double your velocity (just like the elastic
collisions we wer talking about in another thread).
You still need to burn quite a bit of fuel (and
time!) to get into position to make the manouver.
Granted the available boost is significant and free
after that, but it's another example of a fuel versus
time tradeoff in space travel.
That velocity boost *still* won't allow you to
travel in a straight line to Mars. It'll just give
you a segment of a larger ellipse to follow on
your way out. You'll need to burn fuel and aerobrake
at the other end to insert into Mars orbit.
> That is pretty sad ignoring all that energy you
> used to take the longer path.
> You did nto have to store it true, but you still
> needed it.
In space travel, anything you don't have to carry is
free. The only currencly that matters is available
Delta-V = fuel = energy. You can't make a potential
versus kinetic energy trade in space without burning
*some* fuel.
> Again,
> once the slingshot is done.
> The uphill trip to Mars is still there.
> You better have enough energy stored now.
If you did the manouver correctly, you have it stored
as kinetic energy of motion.
> I know a straight path up the hill will need less
> energy than the spiral path you wish to take.
No, then you'll be burning energy (fuel) to try to
follow a straight path, when nature will let you travel
the elliptical one for free.
> It is pretty sad you won't admit that fact at all.
I won't because it's not true.
> Apparently you do have some form of violation
> of the conservation of energy occuring in your
> magical uphill spiral path.
Apparently you can't come to terms with the different
environment for travel that space provides as opposed
to travel on the surface of the Earth where we can
always grab onto the surface to manouver, break for
free, and so forth. In space, any manouver at all
that departs from an inertial path costs fuel. And
fuel is more precious than gold in space.
Spaceman
> No it doesn't. It makes an even trade of potential
> energy for kinetic on the way down, then reverses
> that and trades its kinetic energy for potential
> energy on the way back up.
Gravitatinal potential is energy Greg.
otherwise things that fell far would not have more energy
than things that did not fall as far.
You really should think more about gravity.
> Net change, zero. And
> no engine required to do it either, Nature provides
> the service for free. So no energy expenditure.
Free energy, yes,
but more is needed for longer paths.
You truly are lost on that fact huh?
> The other guy, though, has to burn a whole lot of
> unrecoverable energy resources to take the shorter
> trip.
Not once the inertia is gained.
If he has the same inertia as the long trip taking moron,
he will simply get up that hill faster.
Do you really think a marble rolling at 10 mph will not
roll further up a hill with a straight line than a curved
around the hill path?
:)
You truly are ignorant.
It is very sad you just don't admit you are wrong and
more "actual" energy (no matter where it comes from)
is always needed for a longer path traveled.
At least I know I can be wrong sometimes,
It is sad you know you can't be wrong ever.
Spaceman
> What energy is that? In order to gain energy by
> slingshot manouver around the Earth, you first need
> to spend several months travelling away from the
> Earth and picking up speed to lap the Earth in its
> orbit (thus coming up on it from behind). If you're
> lucky and you do the manouver just right, you can
> almost double your velocity (just like the elastic
> collisions we wer talking about in another thread).
>
> You still need to burn quite a bit of fuel (and
> time!) to get into position to make the manouver.
> Granted the available boost is significant and free
> after that, but it's another example of a fuel versus
> time tradeoff in space travel.
>
> That velocity boost *still* won't allow you to
> travel in a straight line to Mars. It'll just give
> you a segment of a larger ellipse to follow on
> your way out. You'll need to burn fuel and aerobrake
> at the other end to insert into Mars orbit.
WRONG!
If you have enough energy to make it up a hill
spirally to point B then you had enough energy
to travel straight up that hill to point B instead.
You truly have no clue about facts like such.
And I can see you will never admit you are wrong
about it.
Try doing an experiement with marbles or something
Greg,
the spiral path up will not take you as
far up the hill as the straight path will.
go ahead I dare you to try such.
You might learn something from a "Real" experiment you
actually do yourself for once.
:)
<snipped rest of ignorance til Greg rolls the freakin
marble up the hill and admits he is wrong.>
:)
It's easy to set up too Greg,
One ramp for downhill inertia, and
2 setups of paths up the hill,
one a striahgt line from the bottom of your
inertial ramp and one with a curved ramp around
the hill.
both ramps will end at point B on the hill.
Se which one gets there first and at the highest speed
when it gets there.
IF your curved ramp can even do as high as the straight
ramp will!
LOL
Wayne Throop
: You truly are ignorant.
Projection.
: Do you really think a marble rolling at 10 mph will not roll further
: up a hill with a straight line than a curved around the hill path?
If you mean, one goes up a steep slope, and the other a less steep
slope, then indeed it will not roll further up the hill, presuming
there are no friction losses. In orbits in vacuum, there are no
friction losses.
: It is very sad you just don't admit you are wrong and more "actual"
: energy (no matter where it comes from) is always needed for a longer
: path traveled.
As stated, that turns out not to be the case. You get the same energy
by dropping a marble straight down one foot as rolling it down a longer
inclined path the same vertical distance. You get the same energy at
the bottom. Escape velocity, and hence energy, is a scalar, and the
path doesn't matter.
: At least I know I can be wrong sometimes,
And this seems to be one of those times where you are wrong.
Wayne Throop thr...@sheol.org http://sheol.org/throopw
Spaceman
>> "Spaceman" <spac...@yourclockmalfunctioned.duh>
>> Do you really think a marble rolling at 10 mph will not roll further
>> up a hill with a straight line than a curved around the hill path?
>
> If you mean, one goes up a steep slope, and the other a less steep
> slope, then indeed it will not roll further up the hill, presuming
> there are no friction losses. In orbits in vacuum, there are no
> friction losses.
That is not what I was talking about.
Please read that statement again.
The marble has a speed of 10 mph and is moving up the hill.
not down it.
and...
We are not talking orbits here, we are talking
about leaving Earth and traveling to Mars.
AWAY from the sun in other words.
Traveling away from gravity has no drag then?
The Sun has no frictional properties in the form of gravitational drag?
:)
>> It is very sad you just don't admit you are wrong and more "actual"
>> energy (no matter where it comes from) is always needed for a longer
>> path traveled.
>
> As stated, that turns out not to be the case. You get the same energy
> by dropping a marble straight down one foot as rolling it down a
> longer inclined path the same vertical distance. You get the same
> energy at the bottom. Escape velocity, and hence energy, is a
> scalar, and the path doesn't matter.
You seem to have mixed up what I stated.
The path down is giving the energy.
a longer fall will of course give more energy (until terminal velocity
is met anyways)
But that is not what I was talking about.
Try this:
Make a straight ramp that a marble or whatever ball you want
will roll down it.
This will be your inertial energy. (created by gravity of course.)
Now make the ball roll up a curved path around a hill towards
the top of the hill and mark the height it achieved.
Now on that same hill place a straight ramp to the same ending
height that the other hill ramp made it to.
And check to see if it makes it or if it does not, or if it makes
it with energy to spare from not wasting energy traveling
around the spiral ramp.
:)
Then after you do this.
You will find out a straight path uses less energy than a curved
path to the same point does.
If you do not believe this energy fact.
You really need to do the experiement to see such.
:)
and
BTW:
If I am wrong about the short path having more energy left
than the long path marble did then.
There is a problem with conservation of energy and you just proved
CoE wrong.
Greg Neill
> Do you really think a marble rolling at 10 mph will not
> roll further up a hill with a straight line than a curved
> around the hill path?
If there are no frictional losses on the curved path,
it will roll to precisely the same height.
The initial specific kinetic energy in both cases
is the same: (1/2)*v^2, it will be traded for
specific potential energy as it rises in height:
g*dh.
[note to James, "specific" energy is the energy per
unit mass. Since you didn't provide the mass of the
marble, I used the generic specific energies my
description, so (1/2)*v^2 rather than (1/2)*m*v^2
and g*dh rather than m*g*dh. As you should be able
to see, the actual mass of the marble is not required
to be able to compare the energies provided it is
the same in both expressions).
Spaceman
> "Spaceman" <spac...@yourclockmalfunctioned.duh> wrote in message
> news:nZ-dnXMwqN7cu-3V...@comcast.com
>>
>> Do you really think a marble rolling at 10 mph will not
>> roll further up a hill with a straight line than a curved
>> around the hill path?
>
> If there are no frictional losses on the curved path,
> it will roll to precisely the same height.
So that basically should show you,
If there were no Gravity of the sun your longer path is fine.
but guess what..
BBZZZZZZ the Suns gravity is basically a frictional force
that will cause drag when trying to move away from it towards
Mars..
So..
Now say you were wrong about it or twist around even more
because you could never be wrong.
:)
C,mon Greg,
I know you can say it Fonzie..
Greg Neill
You keep changing the scenario from space to hills,
which have some differences. In space there's no
surface to follow to keep you from falling in the
direction of gravity. On a hillside you use the
hills surface to provide that support.
In space, the only way to travel a straight line
in free-fall is to travel radially in the
gravitational field. In the earth to Mars scenario
you would have to travel outward along the line that
passes through the Sun and Earth, at least for the
start of your journey when the Earth's gravity is
a significant factor. Of course, the Earth is in
orbit and won't stay aligned with you for long.
If you choose to travel this radial path, then you
must discard the "free" velocity that you have
already got by simply travelling along with the
Earth in its orbit. That's throwing away about
30 km/sec of speed. Not only that, but you can't
just apply breaks to shed that velocity. You have
to perform an engine burn to do it, a Delta-V of
30 km/sec. That's before you even get started on
your way to Mars. After that you need to expend
the energy to reach the higher potential energy of
Mars' orbit. After *that* you need to expend the
energy to match Mars' speed in its orbit. That's
another 24 km/sec Delta-V.
So I figure you've burnt about 60 Delta-V units
in fuel to drive your straight line. Compare this
with a Delta-V of about 5 or 6 for a Hohmann transfer.
Rocket scientists aren't dummies.
> Try doing an experiement with marbles or something
> Greg,
> the spiral path up will not take you as
> far up the hill as the straight path will.
If there's no friction, it certainly will. You're
just trading kinetic energy for potential energy.
Greg Neill
> than the long path marble did then.
> There is a problem with conservation of energy and you just proved
If you are wrong. They both have the same energy,
and energy is conserved.
Spaceman
Not really,
The Sun is the hill.
<snipped rest of babble based upon not thinking
the Sun's gravity will cause "drag" just as a hill would.>.
Spaceman
They both have the same energy (inertia ramp)
to fight the hill (the Sun's gravity.)
The longer path loses.
More time being "dragged".
Sheesh!
Get a clue about using comparitive thoughts Greg.
Hill = Sun.
You can stay right all you want forever Greg.
It is only proof how wrong you could be and just
won't admit it because of a pride issue.
You are just afraid to actually try the experiment
because it will prove you wrong.
:)
I used to do stuff like this with a tennis ball and sand
when I was a kid.
It seems you must have missed experimenting as a child,
and maybe still are missing it today.
Greg Neill
>> "Spaceman" <spac...@yourclockmalfunctioned.duh> wrote in message
>> news:nZ-dnXMwqN7cu-3V...@comcast.com
>>>
>>> Do you really think a marble rolling at 10 mph will not
>>> roll further up a hill with a straight line than a curved
>>> around the hill path?
>>
>> If there are no frictional losses on the curved path,
>> it will roll to precisely the same height.
>
> So that basically should show you,
> If there were no Gravity of the sun your longer path is fine.
No, it shows that in a gravitational field (the marbles
are rolled in a gravitational field) the path does not
matter for the overall energy expended in moving from
one point to another.
> but guess what..
> BBZZZZZZ the Suns gravity is basically a frictional force
> that will cause drag when trying to move away from it towards
> Mars..
No. Gravity is not a frictional force. It is lossless.
It conserves energy. That's why it's called a conservative
field. What it takes away in potential energy it gives you
back in full measure in kinetic energy, and vice versa.
What you do with yuur fuel on your own time is something
else (like drive in circles or in straight lines); the
field only bothers with changes in height in the field.
And in fact the marble demonstration shows that the path
doesn't matter for total energy change achieved. What does
matter, as far as the path is conserned, is how you fritter
away energy making manouvers off of an inertial trajectory.
> So..
> Now say you were wrong about it or twist around even more
> because you could never be wrong.
I'll say I'm wrong when I'm wrong. Not till then.
Wayne Throop
: Please read that statement again.
OK.
::: Do you really think a marble rolling at 10 mph will not roll further
::: up a hill with a straight line than a curved around the hill path?
Yes, it will not.
: The marble has a speed of 10 mph and is moving up the hill. not down it.
And, in a freely rolling shot, it doesn't matter how long the path,
it reaches the same height.
: Traveling away from gravity has no drag then?
Correct. It has no drag. Traveling through a gas or liquid has drag.
: You seem to have mixed up what I stated.
You stated
::: more "actual" energy (no matter where it comes from) is always needed
::: for a longer path traveled.
and you were incorrect. It's really just that simple.
: If I am wrong about the short path having more energy left than the
: long path marble did then. There is a problem with conservation of
: energy and you just proved CoE wrong.
No, in reality, the fact that the path length doesn't matter is
*required* by conservation of energy.
Greg Neill
>> "Spaceman" <spac...@yourclockmalfunctioned.duh> wrote in message
>> news:wKCdnd7uNeBouu3V...@comcast.com
>>>
>>> WRONG!
>>> If you have enough energy to make it up a hill
>>> spirally to point B then you had enough energy
>>> to travel straight up that hill to point B instead.
>>> You truly have no clue about facts like such.
>>> And I can see you will never admit you are wrong
>>> about it.
>>
>> You keep changing the scenario from space to hills,
>> which have some differences.
>
> Not really,
> The Sun is the hill.
>
> <snipped rest of babble based upon not thinking
> the Sun's gravity will cause "drag" just as a hill would.>.
Saw more damning evidence so you snipped it rather
than deal with it, eh? Typical.
Wayne Throop
: <snipped rest of babble based upon not thinking
: the Sun's gravity will cause "drag" just as a hill would.>.
Whether you claim it's "babble" or not, it's still true.
Gravity causes no drag. Friction causes drag.
Martha Adams
"Robert Clark" <rgrego...@yahoo.com> wrote in message
news:8cf7edc9-74e4-40a2...@m45g2000hsb.googlegroups.com...
The distance from Earth to Mars is about 60,000,000 km at closest
approach. If we have a 30 km/sec initial velocity to Mars, which might
be achievable with airbreathing(scramjet) or nuclear propulsion then
the travel time might be 23 days if you make a simplifying assumption
of a straight-line trip. However, the time required to make the
journey might be made significantly better than this 23 days.
The key fact is that the Earth itself has a 30 km/s velocity around
the Sun that can be used to give us an extra velocity boost toward the
orbit of Mars. In this new estimate I'll simplify the analysis by
assuming that at this high velocity and at the short travel time
achieved, the path will be essentially straight, rather than the
actual ellipse.
The famous Hohmann transfer orbit gives a minimal delta-v and energy
solution for traveling from one orbit to another but this is known for
its long travel times, 6 to 7 months for a Earth to Mars trip. We want
to shorten that for a manned trip to reduce the exposure to radiation
and to reduce the effects of long periods in zero-g.
I'll take the Earth orbit radius to be 150 million km and the Mars
orbit radius to be a little more than its distance at perihelion 210
million km. If we went in a tangential direction to Earth's orbit we
would have a total velocity toward the orbit of Mars of 60 km/s. The
problem here is that we would also have a longer straight-line travel
distance. This would result in the travel time being longer than
moving radially at 30 km/s. So the idea is to move at an optimal angle
that can use the Earth's orbital velocity while at the same time not
making the travel distance too long. See the image here for the
diagrams to illustrate the addition of velocities at an angle θ
(theta) and the travel distance at the angle θ calculations:
http://www.advancedphysics.org/forum/attachment.php?attachmentid=282
(may need to do a free registration at www.advancedphysics.org to
access the image.)
In the diagrams v is the total velocity, r is Earth's orbital radius,
R is Mars orbital radius and d is the straight-line travel distance.
Applying the law of cosines for the velocities gives for the total
velocity:
v^2 = 30^2 + 30^2 -2(30)(30)cos(180-θ) = 2(30^2)(1 + cos(θ))
So v = 30*sqrt(2(1 + cos(θ))
Applying the law of cosines for the travel distance gives the
equation:
R^2 = r^2 + d^2 -2(r)(d)cos(90 + θ) = r^2 + d^2 + 2(r)(d)sin(θ)
Solving for the travel distance d using the quadratic formula gives:
d = -rsin(θ) + sqrt(R^2 -(rcos(θ))^2)
I created a table using various angles θ in fractions of π (pi)
radians to find the shortest trip time:
θ (radians)| time (days)
-----------------------------
0 | 28.36
Ï€/2 | 16.4
Ï€/3 | 11.4
Ï€/4 | 12.9
Ï€/5 | 14.8
Ï€/6 | 16.4
Ï€/7 | 17.7
Ï€/8 | 18.7
Ï€/9 | 19.6
We see the shortest time at π/3 or 60 degrees is a surprising 11.4
days.
Quite a significant advantage than taking a 6 month long Hohmann
transfer orbit.
I'll assume like Robert Zubrin, author of the "Mars Direct" plan,
that aerobraking can be used to stop at Mars on arrival even at such
high speeds.
How reasonable is the assumption that at such high speeds and the
resulting short travel times the straight-line approximation is
accurate?
Bob Clark
===============================================================
For whatever reason, my software didn't mark the preceding
message as what I'm replying to. Anyhow, a couple of
questions occur to me:
1) For the above travel times, when your ship arrives at Mars
orbit, what will be its velocity with respect to Mars; and in what
direction?
2) Can you just outline how ships are to be made, that will
accomplish the short travel times you propose? What
fuel, how much, and etc? ??
3) Where velocity must be shed to land on Mars, how will
your ship do that?
Titeotwawki -- mha [2008 Jly 05]
Spaceman
>> "Spaceman" <spac...@yourclockmalfunctioned.duh>
>> Please read that statement again.
>
> OK.
>
>>>> Do you really think a marble rolling at 10 mph will not roll
>>>> further up a hill with a straight line than a curved around the
>>>> hill path?
>
> Yes, it will not.
Try again.
Better yet,
Try an actual experiment doing such.
:)
>> The marble has a speed of 10 mph and is moving up the hill. not
>> down it.
>
> And, in a freely rolling shot, it doesn't matter how long the path,
> it reaches the same height.
So you never have done such an experiment in real life huh?
LOL
>> Traveling away from gravity has no drag then?
>
> Correct. It has no drag. Traveling through a gas or liquid has drag.
Wrong!
If gravity does not have a "drag" then we could jump off the earth
right through the air and all.
Sheesh.
>>>> more "actual" energy (no matter where it comes from) is always
>>>> needed for a longer path traveled.
>
> and you were incorrect. It's really just that simple.
No,
In fact if I am incorrect about more energy needed for a longer path.
Then the conservation of energy is also wrong.
But I know better than that.
It seems you don't.
LOL
> No, in reality, the fact that the path length doesn't matter is
> *required* by conservation of energy.
Wrong again.
The conservation of energy depends on path length also
when you are dealing with energy use over a distance.
and through a "drag". (friction)
Spaceman
> No. Gravity is not a frictional force.
That is so wrong it again is not funny.
I am forgetting this joke you are pulling on yourself.
Apparently your gravity does not cause friction as in a drag
would as we move upward.
so we could just jump off the Earth.
LOL
Nevermind Greg.
Take the long way all you want.
I will wait for you there.
Spaceman
> "Spaceman" <spac...@yourclockmalfunctioned.duh> wrote in message
> news:E8KdnQudtZdrp-3V...@comcast.com
>> Greg Neill wrote:
>>> "Spaceman" <spac...@yourclockmalfunctioned.duh> wrote in message
>>> news:wKCdnd7uNeBouu3V...@comcast.com
>
>>>>
>>>> WRONG!
>>>> If you have enough energy to make it up a hill
>>>> spirally to point B then you had enough energy
>>>> to travel straight up that hill to point B instead.
>>>> You truly have no clue about facts like such.
>>>> And I can see you will never admit you are wrong
>>>> about it.
>>>
>>> You keep changing the scenario from space to hills,
>>> which have some differences.
>>
>> Not really,
>> The Sun is the hill.
>>
>> <snipped rest of babble based upon not thinking
>> the Sun's gravity will cause "drag" just as a hill would.>.
>
> Saw more damning evidence so you snipped it rather
> than deal with it, eh? Typical.
No,
snipped it because it is yet another one of your
do the experiment this way instead of the way stated.
You are the one that kept removing the experiment
I told you to try.
The hill was the Suns gravity.
You are afraid of the silly little marble experiment.
LOL
Nevermind Greg.
I will be waiting for you up the top of the hill.
and I will of course.. take the shorter route
so I might be waiting for a while.
LOL
Spaceman
>> "Spaceman" <spac...@yourclockmalfunctioned.duh>
>> <snipped rest of babble based upon not thinking
>> the Sun's gravity will cause "drag" just as a hill would.>.
>
> Whether you claim it's "babble" or not, it's still true.
> Gravity causes no drag. Friction causes drag.
So What does gravity cause that stops us from jumping off
Earth?
LOL
Wayne,
If the same amount of energy is needed to get to
the same spot on the hill no matter the path.
Then guess what?
The straight path to Mars will be faster.
and not use anymore energy once the slingshot gives
the needed velocity.
Sheesh.
This is funny,
First.. there was no way the short path could work
at all and now both paths use the same energy.
You both should make up your mind.
LOL
If you take the long way.
I will beat you there and use the same amount of energy
in the end, even according to both of you now.
Simple as that.
LOL
:)
Greg Neill
>> "Spaceman" <spac...@yourclockmalfunctioned.duh> wrote in message
>> news:t9KdnWWuzd04qe3V...@comcast.com
>>> If I am wrong about the short path having more energy left
>>> than the long path marble did then.
>>> There is a problem with conservation of energy and you just proved
>>
>> If you are wrong. They both have the same energy,
>> and energy is conserved.
>
> They both have the same energy (inertia ramp)
> to fight the hill (the Sun's gravity.)
> The longer path loses.
> More time being "dragged".
Both marbles will rise to the same height. Gravity is
a conservative force, it doesn't "steal" energy like
friction.
By your logic, all the planets should have long since
lost all their kinetic energy and fallen into the Sun.
> Sheesh!
> Get a clue about using comparitive thoughts Greg.
>
> Hill = Sun.
>
> You can stay right all you want forever Greg.
> It is only proof how wrong you could be and just
> won't admit it because of a pride issue.
>
> You are just afraid to actually try the experiment
> because it will prove you wrong.
Just about every experiment you've ever suggested has
been done decades or centuries ago in one form or
another, and the results are in the physics laws we
use. Your marble experiments are high school physics
labs, which I did in grade 10 I think. Of course we
had to analyse the kinetic and potential energy states
and draw graphs and perform rudimentaty error analysis
and so forth. A lot more than just "roll a marble and
see what happens".
>
> I used to do stuff like this with a tennis ball and sand
> when I was a kid.
> It seems you must have missed experimenting as a child,
> and maybe still are missing it today.
I built crystal radios, amplifiers and oscillators and
stuff, and various electromechanical machines with small
motors, relays, and Mechano set bits, looked at the planets
and stars with a telescope, prepared microscope slides of
samples I collected, grew various things in petrie dishes
to look at under the scope, tinkered with a chemistry set
(but never really took to it), took apart just about
every toy I ever had to see how it worked, and read anything
and everything that I could get my hands on that dealt with
science. Apparently I was a handfull according to my older
siblings. My parents just encouraged me.
Wayne Throop
: I used to do stuff like this with a tennis ball and sand
: when I was a kid.
Objects that are notoriously subject to friction.
Which objects in space are notoriously not.
Greg Neill
>>> "Spaceman" <spac...@yourclockmalfunctioned.duh>
>
>>> Traveling away from gravity has no drag then?
>>
>> Correct. It has no drag. Traveling through a gas or liquid has
>> drag.
>
> Wrong!
> If gravity does not have a "drag" then we could jump off the earth
> right through the air and all.
Drag is a frictional force. It is lossy, in that energy is
irrecoverably lost to frictional forces.
Gravity is a conservational force, it gives back what
it takes. You can jump off the Earth and into the air
all right, but gravity will insist on converting your
potential energy into kinetic energy as soon as it can.
> Sheesh.
>
>>>>> more "actual" energy (no matter where it comes from) is always
>>>>> needed for a longer path traveled.
>>
>> and you were incorrect. It's really just that simple.
>
> No,
> In fact if I am incorrect about more energy needed for a longer path.
> Then the conservation of energy is also wrong.
> But I know better than that.
> It seems you don't.
> LOL
>
>
>> No, in reality, the fact that the path length doesn't matter is
>> *required* by conservation of energy.
>
> Wrong again.
> The conservation of energy depends on path length also
> when you are dealing with energy use over a distance.
> and through a "drag". (friction)
There's no friction in a conservative field like gravity.
That's why the planets are still orbiting.
Spaceman
> Both marbles will rise to the same height.
Thanks,
That means a straight path to mars will get you there faster
from the same amount of energy once you have enough
inertia to make it there at all.
Case closed you lose.
Admit you are wrong or stay for ever a jackass.
:)
Greg Neill
>> No. Gravity is not a frictional force.
>
> That is so wrong it again is not funny.
> I am forgetting this joke you are pulling on yourself.
> Apparently your gravity does not cause friction as in a drag
> would as we move upward.
> so we could just jump off the Earth.
You could just jump off the Earth if you had the
legs for it. Sadly though, gravity insists that
you first pay it in full for the potential energy
it has on your tab, and your legs just aren't up
to providing that kind of kinetic cash on short
notice (gravity will only take kinetic energy in
payment for potential debt).
> LOL
> Nevermind Greg.
> Take the long way all you want.
> I will wait for you there.
Somehow I feel that you will get lost along the
way, even taking the short route.
Spaceman
>> "Spaceman" <spac...@yourclockmalfunctioned.duh>
>> I used to do stuff like this with a tennis ball and sand
>> when I was a kid.
>
> Objects that are notoriously subject to friction.
> Which objects in space are notoriously not.
So you think the Sun has no "gravitational force" at all?
Greg Neill
> Greg Neill wrote:
>>
>> Saw more damning evidence so you snipped it rather
>> than deal with it, eh? Typical.
>
> No,
> snipped it because it is yet another one of your
> do the experiment this way instead of the way stated.
Perhaps then you could specify precisely the route
you are proposing to take for the journey from Earth
orbit to Mars. Be sure to specify the relative
locations of the Sun, Earth, and Mars. Then we can
work out the energy expenditure details.
Spaceman
> "Spaceman" <spac...@yourclockmalfunctioned.duh> wrote in message
> news:F_idnWADeb8You3V...@comcast.com
>> Wayne Throop wrote:
>>>> "Spaceman" <spac...@yourclockmalfunctioned.duh>
>>>> Traveling away from gravity has no drag then?
>>>
>>> Correct. It has no drag. Traveling through a gas or liquid has
>>> drag.
>>
>> Wrong!
>> If gravity does not have a "drag" then we could jump off the earth
>> right through the air and all.
>
> Drag is a frictional force. It is lossy, in that energy is
> irrecoverably lost to frictional forces.
>
> Gravity is a conservational force, it gives back what
> it takes.
It only give back if you get slowed down enough by it
to cause the reverse motion.
It is like a friction until the friction starts pulling back.
> There's no friction in a conservative field like gravity.
> That's why the planets are still orbiting.
So, no reason a straight line after slingshot event will not work.
Sheesh!
C,mon Greg.
Admit it for once, you were wrong.
:)
Wayne Throop
: LOL
Laughing doesn't demonstrate you right, whether you do it out loud or
titter behind your hand. To anybody who's taken physics and done lab
experiments carefully and knows what friction is, it demonstrates you
aren't listening, and don't know what the words you are uttering mean.
Spaceman
What velocity is needed to "catch" something in
an orbit of Mars?
Spaceman
> Laughing doesn't demonstrate you right, whether you do it out loud or
> titter behind your hand. To anybody who's taken physics and done lab
> experiments carefully and knows what friction is, it demonstrates you
> aren't listening, and don't know what the words you are uttering mean.
No friction OK..
So, no reason a straight line after slingshot event will not work
if timed correctly to meet Mars in its orbital path?
Like shooting skeet without friction of the air and
without gravity pulling so much.
:)
Wayne Throop
: "Spaceman" <spac...@yourclockmalfunctioned.duh>
: That is so wrong it again is not funny.
Half right. It's not funny. That's because it's correct.
Why don't you look up the word "friction" in an encyclopedia,
physics text, or at least a dictionary. You clearly don't know
what the words you are using mean.
: Take the long way all you want. I will wait for you there.
Now you're changing the topic, to elapsed time, not energy expended,
nor height reached on a hill. Sure, now you're back to the topic
thread, but that still doesn't make all your claims about energy,
hills, and gravity correct. Those are still incorrect.
And even then, the fastest *elapsed time* freely falling to Mars starting
from earth orbit is going to be a curve. As are most reasonable powered
trjectories. Constant 1-g-or-higher powered trajectories will look
fairly straight, but teven they won't be.
Spaceman
>>> Gravity is not a frictional force.
>
>> "Spaceman" <spac...@yourclockmalfunctioned.duh>
>> That is so wrong it again is not funny.
>
> Half right. It's not funny. That's because it's correct.
> Why don't you look up the word "friction" in an encyclopedia,
> physics text, or at least a dictionary. You clearly don't know
> what the words you are using mean.
To use a word so exact only shows how anal you really are.
Can you answer the question please.
What uses more energy to get to the top of
the hill, a curved path around and up the hill
or a straight path up the hill?
And remove friction if you want to.
A:) Both the same
B:) The short path
C) The long path.
Mitch Raemsch
> Robert Clark wrote:
> > Â The distance from Earth to Mars is about 60,000,000 km at closest
> > approach. If we have a 30 km/sec initial velocity to Mars, which might
> > be achievable with airbreathing(scramjet) or nuclear propulsion then
> > the travel time might be 23 days if you make a simplifying assumption
> > of a straight-line trip.
>
> when plowing into the surface?
>
> What is the cost of fuel/unit of mass of the craft comparing traditional
> transfer orbits and your 23 "strait shot"?
You are going to use fuel energy to lower Kinetic energy. A net
destruction of energy.
Mitch Reamsch
Virgil
"Spaceman" <spac...@yourclockmalfunctioned.duh> wrote:
> The shortest distance being a straight line will
> always take the least amount of energy,
> and the longer the distance will take more energy
> always.
A brachistochrone takes no more energy and less time than a straight
line.
Wayne Throop
: If the same amount of energy is needed to get to the same spot on the
: hill no matter the path. Then guess what? The straight path to Mars
: will be faster.
You guessed wrong. The fastest freely falling trajectory to mars
is going to be curved. In fact, it works better if you don't guess,
and actually try calculating or simulating the cases.
: First.. there was no way the short path could work at all
You of course ahve a reference to where I said a short path "will not
work at all". Please refresh my memory, cupplying a cite, since near
as I can tell, I never said that, first, last, or in between.
A "short path" or "stright path" can, in theory, be made to work.
With vast excess and impractical expenditures of delta-v. Trajectories
starting at rest wrt earth and ending at rest wrt mars which take least
delta-v are curved. If the delta-v is high enough, it can be made to
look fairly straight, but it will be curved.
Greg Neill
>> Both marbles will rise to the same height.
>
> Thanks,
> That means a straight path to mars will get you there faster
> from the same amount of energy once you have enough
> inertia to make it there at all.
Not the same amount of energy given the real life
circumstances, as I tried to explain to you before
but you just snipped and ignored it.
Yes, the potential energy "hill" from Earth's location
to Mars' location is the same either way. So the
same amount of kinetic energy must be provided by
fuel to supply the potential difference.
HOWEVER. The story does not end there. The planets
are not stationary points that you're travelling
between. They are in orbit and going at respectable
speeds, particularly compared to what we can do with
current spacecraft.
In order to travel a straight line path from the
"height" of the Earth's orbit to the "height" of
Mars' orbit you either have to discard the Earth's
orbital velocity that your ship starts with and then
proceed radially outward, or constantly run your
engines to counter the componment of the Sun's
gravitational acceleration that is drawing you
sideways from your desired path. That fuel being
burnt does not get you closer to Mars, it only
maintains your straight line against nature's
desire to have you move along an inertial path.
When you get to the vicinity of Mars, you then have
to match speeds with Mars. Mars most certainly does
not travel in a straight line, so you'll need to
perform what is essentially a right angle turn and
speed up to Mars. That means throwing away all
your outward veloctity and burning fuel to provide an
"orbitward" velocity.
It would be a bit like wanting to take your straight
path up a hill, but you're launching yourself from
a car that's circling the base of the hill at a
high speed. In order to take the straight path you
first have to jump off the car and come to a stop
at the bottom of the hill. Then suppose at the
top of the hill there's a car running around a track
at high speed. You want to jump into that car. So
you then have to get yourself up to the speed of that
car before you can climb aboard.
> Case closed you lose.
> Admit you are wrong or stay for ever a jackass.
I'll let you know when I'm wrong.
Spaceman
Why would an object that has a speed of say 10 mps (energy) take
a shorter time to travel A to B on that curve than a straight line path
would take
at the same speed from A to B?
Do you know what a higher distance means for the same speed?
It means more time to travel it.
You are lost man!
Your brachistochrone is flawed drastically.
Toss it in the scifi math folder.
Wayne Throop
: "Spaceman" <spac...@yourclockmalfunctioned.duh>
: That means a straight path to mars will get you there faster from the
: same amount of energy once you have enough inertia to make it there at
: all.
Nope. Vocabulary tip: I'm going to assume you mean "momentum"
when you say "inertia", and that you are talking about a freely falling
trajectory.
The only way to get a straight free-fall trajectory is to proceed
radially straight out wrt the sun. To do that, you have to "cut across"
the tangential motions of earth and mars. You'll end up taking longer
for just about the same reason that it's faster to cross a moving river
at an angle. Not quite the same issue, but a similar one.
Spaceman
>> "Spaceman" <spac...@yourclockmalfunctioned.duh>
>> If the same amount of energy is needed to get to the same spot on the
>> hill no matter the path. Then guess what? The straight path to Mars
>> will be faster.
>
> You guessed wrong. The fastest freely falling trajectory to mars
> is going to be curved. In fact, it works better if you don't guess,
> and actually try calculating or simulating the cases.
sheesh man,
you don't "free fall" to mars.
You could "freefall" to Venus though.
You have to go "uphill" to mars..
The sun is laughing at you and your "freefall" to Mars.
It is sorta like freefalling up from Earth.
:)
So
Lets try again.
Wayne Throop
: So you think the Sun has no "gravitational force" at all?
No, I simply know what friction is, as you do not.
Try a dictionary, an encyclopedia, or a physics text.
Wayne Throop
:: to take for the journey from Earth orbit to Mars.
: "Spaceman" <spac...@yourclockmalfunctioned.duh>
: What velocity is needed to "catch" something in
: an orbit of Mars?
Aaaaaaand that would be a "no, he can't".
John Park
sci.math dropped from follow-up list
Robert Clark (rgrego...@yahoo.com) writes:
> The distance from Earth to Mars is about 60,000,000 km at closest
> approach.
I believe this is exceptional--the average is more like 77 million km.
If we have a 30 km/sec initial velocity to Mars, which might
> be achievable with airbreathing(scramjet) or nuclear propulsion then
> the travel time might be 23 days if you make a simplifying assumption
> of a straight-line trip. However, the time required to make the
> journey might be made significantly better than this 23 days.
> The key fact is that the Earth itself has a 30 km/s velocity around
> the Sun that can be used to give us an extra velocity boost toward the
> orbit of Mars. In this new estimate I'll simplify the analysis by
> assuming that at this high velocity and at the short travel time
> achieved, the path will be essentially straight, rather than the
> actual ellipse.
[...]
> I'll take the Earth orbit radius to be 150 million km and the Mars
> orbit radius to be a little more than its distance at perihelion 210
> million km. If we went in a tangential direction to Earth's orbit we
> would have a total velocity toward the orbit of Mars of 60 km/s. The
> problem here is that we would also have a longer straight-line travel
> distance. This would result in the travel time being longer than
> moving radially at 30 km/s. So the idea is to move at an optimal angle
> that can use the Earth's orbital velocity while at the same time not
> making the travel distance too long. See the image here for the
> diagrams to illustrate the addition of velocities at an angle =E8
> (theta) and the travel distance at the angle =E8 calculations:
[...]>
>
> We see the shortest time at =F0/3 or 60 degrees is a surprising 11.4
> days.
[...]
> I'll assume like Robert Zubrin, author of the "Mars Direct" plan,
> that aerobraking can be used to stop at Mars on arrival even at such
> high speeds.
I share the reservations about the practicality of all this, but . . .
> How reasonable is the assumption that at such high speeds and the
> resulting short travel times the straight-line approximation is
> accurate?
>
I think in the ball park of 10%. You can get a lower bound on the precision
from conservation of energy. The straight-line approxmation is equivalent
to assuming the velocity is unchanged in th course of the voyage. You know
the kinetic energy and potential energy in Earth orbit (neglect the effect
of the Earth's gravity); you know the potential energy at the position of
Mars' orbit, so you can get the kinetic energy there and hence the speed. I
get a reduction of about 10% from the intial speed.
The change in velocity (i.e. including direction) could well be greater,
but you could put upper bounds on that by assuming the sun's gravitational
acceleration is constant over the voyage, and express the distance in terms
of x and y coordinates. Each will vary as u*t - 0.5*a*t*t, where u and a
are velocity and acceleration components in the x or y direction. u*t is your
straight-line approximation for either component.
For an order-of-magnitude estimate of the correction terms, the gravitational
(centripetal) acceleration due to the sun is (30 km/s)^2/1 AU or about
6*10^-6 km/s. Your travel time is about two weeks or 1.3*10^6 s so
0.5*a*t^2 is about 5 million km, or around 10% of the total distance. It
would be more useful to work out these approximate maximum errors for the x
and y components separately, to give an idea of the possible change
in direction, but I don't imagine the picture will change dramatically.
--John Park
Greg Neill
>> "Spaceman" <spac...@yourclockmalfunctioned.duh> wrote in message
>> news:F_idnWADeb8You3V...@comcast.com
>>> Wayne Throop wrote:
>>>>> "Spaceman" <spac...@yourclockmalfunctioned.duh>
>>>>> Traveling away from gravity has no drag then?
>>>>
>>>> Correct. It has no drag. Traveling through a gas or liquid has
>>>> drag.
>>>
>>> Wrong!
>>> If gravity does not have a "drag" then we could jump off the earth
>>> right through the air and all.
>>
>> Drag is a frictional force. It is lossy, in that energy is
>> irrecoverably lost to frictional forces.
>>
>> Gravity is a conservational force, it gives back what
>> it takes.
>
> It only give back if you get slowed down enough by it
> to cause the reverse motion.
You can also stop for a while (if you've got a place
to stop that'll keep you from falling) and continue
on later. The field "remembers" your potential energy
no matter what.
> It is like a friction until the friction starts pulling back.
The energy is not lost though. It's goes into potential
energy that you will get back as kinetic energy again.
Friction is not so benificent.
>> There's no friction in a conservative field like gravity.
>> That's why the planets are still orbiting.
>
> So, no reason a straight line after slingshot event will not work.
Even after a slingshot maneuver, if you don't constantly
provide more energy by burning fuel then your path will
be a conic section (elliptical, hyperbolic, parabolic...etc.,
depending upon the total mechanical energy). To achieve a
degenerate conic (a radial straight line outwards from the
Sun) via a slingshot maneuver alone is not practical. The
required Delta-V would be too high. You can only double your
speed at best by a slingshot maneuver (that's the theoretical
limit) and since you're starting initially from Earth orbit,
you'll need to first build up a speed difference that you can
double (by burning fuel), then you want the result to cancel
your velocity component Earth-orbitward (about 30 km/sec) and
add the required radial component.
That's way more energy than a simple transfer orbit.
I'm not saying that slingshot maneuvers are not useful.
They can be used to boost a spacecraft into a transfer
orbit via repeated passes over time, so you can save
maybe half the fuel that a straight transfer by engine
insertion would take. But the main portion of the
journey is still going to consist of sections of elliptical
free-fall paths. No one can put enough fuel on a spacecraft
to "drive" it all the way to Mars.
> Sheesh!
> C,mon Greg.
> Admit it for once, you were wrong.
Nope. Not this time.
Spaceman
> "Spaceman" <spac...@yourclockmalfunctioned.duh> wrote in message
> news:HbGdnXSziq-S2e3V...@comcast.com
>> Greg Neill wrote:
>>> Both marbles will rise to the same height.
>>
>> Thanks,
>> That means a straight path to mars will get you there faster
>> from the same amount of energy once you have enough
>> inertia to make it there at all.
>
> Not the same amount of energy given the real life
> circumstances, as I tried to explain to you before
> but you just snipped and ignored it.
>
> Yes, the potential energy "hill" from Earth's location
> to Mars' location is the same either way. So the
> same amount of kinetic energy must be provided by
> fuel to supply the potential difference.
So again,
the straight path once slingshot is achieved use no more
energy in the path itself.
Admit you were wrong on that part at least Greg.
I know you need more energy to slow it down.
but you stated you need more energy just to
take the straight path itself.
> HOWEVER. The story does not end there. The planets
> are not stationary points that you're travelling
> between. They are in orbit and going at respectable
> speeds, particularly compared to what we can do with
> current spacecraft.
Correct,
you have to aim and shoot,
Like shooting ducks that fly in circles without the friction
and very little gravity compared to here on Earth.
:)
> In order to travel a straight line path from the
> "height" of the Earth's orbit to the "height" of
> Mars' orbit you either have to discard the Earth's
> orbital velocity that your ship starts with and then
> proceed radially outward, or constantly run your
> engines to counter the componment of the Sun's
> gravitational acceleration that is drawing you
> sideways from your desired path.
Greg a straight shot from Earth to Mars is a radial path
away from the Sun. The sun pulls from behind.
Your path is the side pulled path.
> That fuel being
> burnt does not get you closer to Mars, it only
> maintains your straight line against nature's
> desire to have you move along an inertial path.
I need no maintaining path, I only need maintained
inertia away from the Sun in the radial path that
a straight shot would be.
Why are you now confusing things your path would
need with things my path would need?
Why don't you just admit you were Wrong about
the straight path energy needed Greg.
The onyl extra energy needed will be to slow
the craft so it does not crash and burn.
(Sadly like Sam said long ago)
> When you get to the vicinity of Mars, you then have
> to match speeds with Mars. Mars most certainly does
> not travel in a straight line, so you'll need to
> perform what is essentially a right angle turn and
> speed up to Mars. That means throwing away all
> your outward veloctity and burning fuel to provide an
> "orbitward" velocity.
correct,
but no extra energy is needed to travel striahgt outward
than would be needed to keep correcting
the suns pull off path in your orbital path.
Again,
If you can get enough velocity to do the orbital path,
I have plenty (actually too much) when I get to Mars.
so yes.. I will need to burn fuel again,
But I will be there much faster than you will.
and .. no extra energy was burned for the trip itself
like I was saying and you still won't admit you were
wrong about such.
>> Case closed you lose.
>> Admit you are wrong or stay for ever a jackass.
>
> I'll let you know when I'm wrong.
You already have been and you are just adding
energy we were not even discussing yet now.
You had stated the straight line path alone would need
more energy to fight the Suns gravity.
You were wrong.
I admit , I will need more nergy to slow down and take
the turn to orbit.
:)
So, your turn to admit you were wrong about
the energy needed to "get there" at all using a straight
path.
C,mon Greg,
Be a Man.
Virgil
"Spaceman" <spac...@yourclockmalfunctioned.duh> wrote:
Absent friction, the energy required is the difference in potential
energies at the two positions, which is independent of the path taken in
conservative fields like the earth's gravitational field.
So unless frictional resistances need to be overcome, the answer is that
they are both the same.
Virgil
"Spaceman" <spac...@yourclockmalfunctioned.duh> wrote:
> Virgil wrote:
> > In article <K6udnYGWU6KFk-3V...@comcast.com>,
> > "Spaceman" <spac...@yourclockmalfunctioned.duh> wrote:
> >
> >
> >> The shortest distance being a straight line will
> >> always take the least amount of energy,
> >> and the longer the distance will take more energy
> >> always.
> >
> > A brachistochrone takes no more energy and less time than a straight
> > line.
>
> Why would an object that has a speed of say 10 mps (energy) take
> a shorter time to travel A to B on that curve than a straight line path
> would take
> at the same speed from A to B?
>
> Do you know what a higher distance means for the same speed?
But I am not going by mere speed, but by the necessary energy
expenditures, and for any given energy expenditure which is sufficient
to get to one's goal at all, there is a brachistochrone which takes
neither more energy nor more time.
> Toss it in the scifi math folder.
I took it out of the real physics folder, where it belongs.
Spaceman
>>> Both marbles will rise to the same height.
>
>> "Spaceman" <spac...@yourclockmalfunctioned.duh>
>> That means a straight path to mars will get you there faster from the
>> same amount of energy once you have enough inertia to make it there
>> at all.
>
> Nope. Vocabulary tip: I'm going to assume you mean "momentum"
> when you say "inertia", and that you are talking about a freely
> falling trajectory.
Freefalling trajectory is not what is occuring on the trip
to Mars.
It is actually a fight for flight until Mars gravity is stronger than
the Sun.
The Sun is the other way.
No freefalling occuring in this path.
Sheesh.
> The only way to get a straight free-fall trajectory is to proceed
> radially straight out wrt the sun. To do that, you have to "cut
> across" the tangential motions of earth and mars. You'll end up
> taking longer for just about the same reason that it's faster to
> cross a moving river at an angle. Not quite the same issue, but a
> similar one.
So you think outerspace is a river flowing?
:)
Shorter paths at same velocities never take longer
than a longer path would..
You really should rethink that.
and
BTW: It is never shorter to cross the river at an angle.
You see walking down into the water is yet anothe curve
added that I don't go for.
I use a bridge and every time that bridge goes straight across
I can cross it faster than the angled bridge if moving at the same speed
in either case.
Now,
If there is a flowing aether in outerspace, you may have a real good point.
And I would have to agree with you then.
Until you prove that, (I have tried.. it is very hard to prove)
I will stick with the straight line being the shortest distance and time
at a certain speed
Wayne Throop
: So, no reason a straight line after slingshot event will not work
: if timed correctly to meet Mars in its orbital path?
What do you mean by "work"? Can it be done? Yes.
Is it going to be the fastest route for a given slingshot? No.
: Like shooting skeet without friction of the air and
: without gravity pulling so much.
Say you are target shooting on a moving train. The target is above
you, moving in the same direction as the train, at a slightly different
velocity. If you wait for the target to be in the right position, you
can shoot with a trajectory straight-wrt-the-track to reach the target.
It won't be the fastest path. If you simply shoot to reach the target
when it's directly above you at minimum distance, you'll get the shortest
path, and the path will be curved wrt the track.
The case for earth-to-mars is more complicated, because inverse-square
becomes significant at that distance, and the relative velocity of earth
and mars is fairly high, and if you want an intact spacecraft or probe
at the end of the trip (ie, no lithobraking), you need to figure on two
"slingshot events" (ie, you have to use some of your delta-v near earth,
and some near mars), but the bottom line is, the fastest trip for a given
total delta-v is going to have a curved path wrt the sun.
Virgil
"Spaceman" <spac...@yourclockmalfunctioned.duh> wrote:
> So
> Lets try again.
> What uses more energy to get to the top of
> the hill, a curved path around and up the hill
> or a straight path up the hill?
> And remove friction if you want to.
>
> A:) Both the same
> B:) The short path
> C) The long path.
Absent friction, both take the same amount of energy.
Spaceman
>> "Spaceman" <spac...@yourclockmalfunctioned.duh>
>
> Try a dictionary, an encyclopedia, or a physics text.
Again,
It is great you use such twists instead of just admitting you
were wrong about a straight path being shorter and taking
no more energy to do such.
I do know I will need more energy to land.
but not more energy to get there if I use the straighest path possible.
:)
It is amazing that not one math person can ever be wrong.
LOL.
:)
Spaceman
>>> Perhaps then you could specify precisely the route you are proposing
>>> to take for the journey from Earth orbit to Mars.
>
>> "Spaceman" <spac...@yourclockmalfunctioned.duh>
>> What velocity is needed to "catch" something in
>> an orbit of Mars?
>
> Aaaaaaand that would be a "no, he can't".
No,
That is a simple question that I would like answered
by what Greg thinks would be needed.
You don't want to answer it either huh?
Wayne Throop
: What uses more energy to get to the top of
: the hill, a curved path around and up the hill
: or a straight path up the hill?
: And remove friction if you want to.
No matter how many times you say it, it still uses the same amount
of *energy*. In this particular case, the shorter path will take less
time for the same energy, but no matter how many times you follow up by
saying "aha, that proves I'm right", it doesn't, and you're still wrong.
Because a trajectory confined to a frictionless surface is not the same
problem as a ballistic (ie, freely falling) trajectory. Comparing a case
of a ballistic trajectory between a moving shooter and a moving target,
and the result is, the best trajectory is curved. Boost the velocity
high enough, and it'll start to look fairly straight, but it'll still
be curved.
Spaceman
> Even after a slingshot maneuver, if you don't constantly
> provide more energy by burning fuel then your path will
> be a conic section (elliptical, hyperbolic, parabolic...etc.,
> depending upon the total mechanical energy).
Apparently you forget what straight meaans again Greg.
Oh sheesh.
If you don't need fuel for your curved path,
I don't need fuel for my straight path.
I only need fuel to not crash into Mars when I am there
faster than you will be.
C,mon Greg.
Just admit you were wrong about the straight line
needing more fuel just to get there.
We proved such with the simple hill problem.
Sheesh.
Greg Neill
>> "Spaceman" <spac...@yourclockmalfunctioned.duh> wrote in message
>>> Greg Neill wrote:
>>
>>>>
>>>> Saw more damning evidence so you snipped it rather
>>>> than deal with it, eh? Typical.
>>>
>>> No,
>>> snipped it because it is yet another one of your
>>> do the experiment this way instead of the way stated.
>>
>> you are proposing to take for the journey from Earth
>> locations of the Sun, Earth, and Mars. Then we can
>> work out the energy expenditure details.
>
> an orbit of Mars?
I'm not sure I understand what you mean. Do you mean,
what's the highest relative velocity you can have
with respect to Mars in order to be able to slow into
orbit by a reasonable maneuver?
First you need to be just about matching Mars in
direction and speed in its orbit (about 24 km/sec).
If you get the approach right, you will line up so
that you just have to shed a bit of excess speed to
stick in an orbit about the planet.
Orbital velocity around Mars depends upon the radius
of the orbit, as does escape velocity. You want to
be below escape velocity. For a radius r around Mars,
the escape velocity is
v_esc = sqrt(2 * 4.283 x 10^4 km^3/sec^2 / r)
r is in km.
At 450km above Mars' surface, the escape speed is about
4.7 km/sec.
So there's a starting point. Get to that height with
just less than that relative velocity and your in.
Spaceman
>> "Spaceman" <spac...@yourclockmalfunctioned.duh>
>> if timed correctly to meet Mars in its orbital path?
>
> Is it going to be the fastest route for a given slingshot? No.
You are lost then if you think it would not.
You actually think the shortest distance between two
points is a straight line don't you?
>> Like shooting skeet without friction of the air and
>> without gravity pulling so much.
>
> Say you are target shooting on a moving train. The target is above
> you, moving in the same direction as the train, at a slightly
> different velocity. If you wait for the target to be in the right
> position, you can shoot with a trajectory straight-wrt-the-track to
> reach the target. It won't be the fastest path. If you simply shoot
> to reach the target when it's directly above you at minimum distance,
> you'll get the shortest path, and the path will be curved wrt the
> track.
You are totally lost..
It is a radial shot and if shot correctly it will be a straight line
between the two points and will be the shortest path and time
to take between the two points.
Observers from other points of view will also see it as a straight
line between the two points if they mark the two points as they pass by.
First mark the shooting point anf then mark the hit point.
It will show as a straight line.
Sheesh
Spaceman
Correct, so a straight line path will take no more energy
to take than a curved path would. except for the fact of entering
Mars orbit itself.
But I never said the total energy to land and all would not be different,
only the energy to get there in a straight line would be no more than the
orbital
path would be.
But sadly nobody gets it.
:(
Wayne Throop
:: line.
: "Spaceman" <spac...@yourclockmalfunctioned.duh>
: Why would an object that has a speed of say 10 mps (energy) take a
: shorter time to travel A to B on that curve than a straight line path
: would take at the same speed from A to B?
Simple. Because if you start the slope out shallow, and only make it
steep nearer the target, you spend more of the trip at a higher speed.
With a straight line, it's losing energy all the way, rather than
preferentially near the end.
If you want to try it with marbles and a little rubber band launcher
calibrated to give a constant starting velocity, you can google images
of brachihstochrone-shaped ramps. Build one and try it.
Of course, that's not the same problem as a ballistic trajectory, but
it does demonstrate the least-elapsed-time distance is not always a
straight line. There are other demonstrations of this in optics and such.
Use google. Read more about it. There's also these big buildings
with lots of books in 'em. "Libraries", I think they're called.
Wayne Throop
: sheesh man,
: you don't "free fall" to mars.
Ah. So you don't know what freefall means either. Shrug.
Try looking up the word.
http://en.wikipedia.org/wiki/Free-fall
Free fall is motion with no acceleration other than that provided by
gravity. Since this definition does not specify velocity, it also
applies to objects initially moving upward.
If you don't like wikipedia, you can look it up elsewhere,
with similar results.
Spaceman
But the whole thing here is a speed already present.
and we are using that speed for the energy,
so at that stated energy (speed) the shorter path will take less time
and no more energy to do such.
>> Toss it in the scifi math folder.
>
> I took it out of the real physics folder, where it belongs.
Yes, but it only applies to frictionless paths with gravity
in one direction only.
(the curve must be convex to the gravity source)
It is a fun little line and actually a cool experiment
I take what I said back it does belong in physics,
I am sorry.
I jumped the gun on it.
:(
Spaceman
>> "Spaceman" <spac...@yourclockmalfunctioned.duh>
>> What uses more energy to get to the top of
>> the hill, a curved path around and up the hill
>> or a straight path up the hill?
>> And remove friction if you want to.
>
> No matter how many times you say it, it still uses the same amount
> of *energy*.
Thanks,
That means the straight line take no more energy to
travel to Mars.
Case closed for a straight line needeing more energy to
take.
It simply won't need such.
Only the orbit course change and landing will take more energy
This is all I said originally and I did goof by saying
fuel instead of energy originally.
Wayne Throop
: So you think outerspace is a river flowing?
No. Do you think it's a hill?
: Shorter paths at same velocities never take longer than a longer path
: would.
And yet again you completely ignore that your starting velocity is changed
as you go along, depending on your path. If you kept the same velocity
the entire path, you'd be right. Since you don't, you'd be wrong.
: BTW: It is never shorter to cross the river at an angle.
It is never a shorter distance. It is always a shorter time.
Find a river and try it. Time yourself.