turn rate vs. airspeed
hsaw...@my-deja.com
(in degrees per second) vary with airspeed? An explanation would be
appreciated.
thanks
Sent via Deja.com http://www.deja.com/
Before you buy.
Roy Smith
> In a level (constant-altitude) turn at a given bank angle, does the turn
> rate (in degrees per second) vary with airspeed? An explanation would be
> appreciated.
Turn rate decreases (and turn radius increases) as airspeed increases. If you
know vector calculus, it's about a 4-line derivation to show why this is so, but
I'm going to assume you're looking for a simplier explanation.
For a given bank angle, you generate a certain amount of force pushing the plane
sideways, changing its direction of flight. As the plane goes faster, it has
more momentum, so the same amount of sideways force will have less of an effect
on its direction of flight, since it has to overcome more momentum.
Now, that's not a rigorously correct explanation from the physics point of view,
but it's intuitive and will get you the right answer.
BTW, this is a fairly easy thing to demonstrate for yourself. Pick a smallish
feature on the ground that's easy to see, such as a lake, field, or whatever.
Slow the plane down to a fairly low speed (final approach speed would be a good
place to start) and roll in 30 degrees of bank, with coordinated rudder and make
a 360 degree turn. Observe how large a circle you make on the ground (this
works best on a day with little or no wind). Also, time how long it takes you
to make a 360 heading change. If the circle you make is substantially larger or
smaller than the ground feature you picked, find another one that's closer to
the side of the circle you're making. Discuss with your instructor what
altitude to do this at, but 1000 AGL should be about right.
Now, get the plane going at cruise speed (in most light planes, that will
typically be about twice the speed you used before), and roll in the same 30
degree bank angle, again with coordinated rudder. Compare the size of the
circle you make and the amount of time it takes to do 360 degrees of heading
change with what you got before.
--
Roy Smith <r...@popmail.med.nyu.edu>
CP-ASEL-IA, CFI-ASE-IA
Tracy R Reed
>degree bank angle, again with coordinated rudder. Compare the size of the
>circle you make and the amount of time it takes to do 360 degrees of heading
>change with what you got before.
So, the faster you go, the longer it will take you to do a 360 degree turn.
This implies that a "standard rate turn" is not the same in every aircraft,
correct? I have not yet begun my instrument training yet but I have been told
that 2 minute turns are the thing to do in IMC and my little 152 seems happy
enough doing a 2 minute 360 degree turn. How long would a 747 take to do a 360?
--
Tracy Reed http://www.ultraviolet.org
In 1998 SCO called Linux a toy operating system. In 2000 a Linux company buys
SCO. Who's next?
John Lowry
Turn rate (in radians per second) = g(in ft/sec^2)*tangent(bank angle)/true
air speed (in ft/sec). A radian is about 57.3 degrees.
--
John T. Lowry, PhD
Flight Physics; 724 Alderson Ave.; Billings MT 59101
Voice: 406-248-2606
Web site: http://www.mcn.net/~jlowry
E-mail: jlo...@mcn.net
<hsaw...@my-deja.com> wrote in message news:8ns0i4$5av$1...@nnrp1.deja.com...
> In a level (constant-altitude) turn at a given bank angle, does the turn
rate
> (in degrees per second) vary with airspeed? An explanation would be
> appreciated.
>
Roy Smith
> So, the faster you go, the longer it will take you to do a 360 degree turn.
For a given bank angle, yes, that is correct.
> This implies that a "standard rate turn" is not the same in every aircraft,
> correct?
A standard rate turn (what our British friends would call a "rate one turn")
takes 2 minutes to complete a 360 heading change, regardless of aircraft type.
For a given airspeed, there will be a specific bank angle which will produce a
standard rate turn. The faster you go, the higher the bank angle has to be to
get standard rate.
> How long would a 747 take to do a 360?
It depends (only) on how fast it is going and what bank angle it is using.
Typically jets limit their bank angle to 25 degrees or less, which is not enough
to produce a standard rate turn at the speed they cruise at (usually about 4-500
KTAS).
Robert Frischer
the airspeed is. The faster plane will have a larger turning radius to do
it. Standard rate turns are done with the turn coordinator.
--
Robert Frischer, rfri...@iname.com
"Tracy R Reed" <tr...@freeside.ultraviolet.org> wrote in message
news:slrn8q348j...@freeside.ultraviolet.org...
> Roy Smith <r...@popmail.med.nyu.edu> wrote:
> >degree bank angle, again with coordinated rudder. Compare the size of
the
> >circle you make and the amount of time it takes to do 360 degrees of
heading
> >change with what you got before.
>
> So, the faster you go, the longer it will take you to do a 360 degree
turn.
> This implies that a "standard rate turn" is not the same in every
aircraft,
> correct? I have not yet begun my instrument training yet but I have been
told
> that 2 minute turns are the thing to do in IMC and my little 152 seems
happy
> enough doing a 2 minute 360 degree turn. How long would a 747 take to do a
360?
>
hsaw...@my-deja.com
In article <wmgo5.9416$p5.3...@nntp1.onemain.com>,
hsaw...@my-deja.com
reverse-engineer it since the "answer" (equation for turn rate as a function
of gravitational acceleration, bank angle, and true air speed) was posted by
someone else in this thread.
I look forward to trying out your suggested exercise, and comparing the
results to the equation's predictions.
In article <roy-09BA2E.1...@netnews.nyu.edu>,
Roy Smith <r...@popmail.med.nyu.edu> wrote:
> hsaw...@my-deja.com wrote:
> > In a level (constant-altitude) turn at a given bank angle, does the turn
> > rate (in degrees per second) vary with airspeed? An explanation would be
> > appreciated.
>
> Turn rate decreases (and turn radius increases) as airspeed increases. If you
> know vector calculus, it's about a 4-line derivation to show why this is so, but
> I'm going to assume you're looking for a simplier explanation.
>
> For a given bank angle, you generate a certain amount of force pushing the plane
> sideways, changing its direction of flight. As the plane goes faster, it has
> more momentum, so the same amount of sideways force will have less of an effect
> on its direction of flight, since it has to overcome more momentum.
>
> Now, that's not a rigorously correct explanation from the physics point of view,
> but it's intuitive and will get you the right answer.
>
> BTW, this is a fairly easy thing to demonstrate for yourself. Pick a smallish
> feature on the ground that's easy to see, such as a lake, field, or whatever.
> Slow the plane down to a fairly low speed (final approach speed would be a good
> place to start) and roll in 30 degrees of bank, with coordinated rudder and make
> a 360 degree turn. Observe how large a circle you make on the ground (this
> works best on a day with little or no wind). Also, time how long it takes you
> to make a 360 heading change. If the circle you make is substantially larger or
> smaller than the ground feature you picked, find another one that's closer to
> the side of the circle you're making. Discuss with your instructor what
> altitude to do this at, but 1000 AGL should be about right.
>
> Now, get the plane going at cruise speed (in most light planes, that will
> typically be about twice the speed you used before), and roll in the same 30
> degree bank angle, again with coordinated rudder. Compare the size of the
> circle you make and the amount of time it takes to do 360 degrees of heading
> change with what you got before.
>
> --
> Roy Smith <r...@popmail.med.nyu.edu>
> CP-ASEL-IA, CFI-ASE-IA
>
>
touch...@my-deja.com
me.....
Take 1/10 of the airspeed and add 5 to get .......?
Seems like it was the apx. bank angle necessary for a standard rate
turn? Or was it the number of degrees to allow for rollout on a steep
turn (45 degrees bank turn)?
Cruise is about 90kts in the 150 I trained in, so the answer is 14.
So....what's the question? (Guess I shoulda written it down).
Touch
Roy Smith, CFI
> Thanks! I wouldn't mind seeing the derivation, although I could probably
> reverse-engineer it since the "answer" (equation for turn rate as a
> function of gravitational acceleration, bank angle, and true air speed) was posted
> by someone else in this thread.
How about I give you some hints how to go about it, and not spoil all
the fun :-)
Start with a banked airplane and decompose the lift vector into
horizontal and vertical components. The vertical component has to equal
the weight of the plane (you'll get the load factor out of this too).
Now, draw a picture looking down. Draw a circle with the airplane on
it. The airplane's velocity vector is tangent to this circle. The
horizontal component of lift that you found in step one points in along
a radius of the circle. If you write F = mA (where F and A are
vectors), you'll see that F is the horizontal component of lift, and A
is just dV. You've got a differential triangle with one side V, and the
perpendicular side dV. The angle opposite dV is d(theta). d(theta)/dt
is the rate of turn. Once you've got rate of turn, observe that the
circumference of the circle is 2*pi*R, and since you know how long it
takes the plane to make a circle (from the rate of turn), and you know
its velocity, you know how far it advances in that time. Equate that to
2*pi*R and solve for R, the radius of turn.
I think this is a particularly elegant physics problem, because it turns
out all you need is basic geometry and first semester calculus (maybe
not even) to solve what at first glance looks like a very complicated
situation.
It's also kind of neat to observe that mass drops out of the equation
entirely. And nowhere do you find any mention of any aerodynamic
qualities of the aircraft. This is why when comparing the turn
performance of the 152 with the 747, all that mattered was bank angle
and speed; if you could get a 747 to fly as slow as a 152 without
stalling, its turn performance would be identical for the same bank
angles.
Of course, what is left out is that some combinations of speed and rate
of turn will imply load factors large enough to tear the wings off the
airplane. This is simply an implemention detail and you should not let
it detract from the theoretical elegance of the equations. Or, to put
it in computer terms, "That's a hardware problem" :-)
--
Roy Smith, CFI-ASE-IA
m c wallace
Roy Smith <r...@popmail.med.nyu.edu> wrote in message
news:roy-09BA2E.1...@netnews.nyu.edu...
around a point I'll try and incorporate this explaination into my lesson
plans. A practical application of this has to to with approach and landing
at high altitude airports. We have several here in So.Cal. and trying to
get students to learn to compensate for this characteristic when coming from
a sea level airport and landing, for example, at Big Bear 6200+/_ can
sometimes be difficult for students to understand. Thanks for the tools.
> -- R.Wallace CFIAIM
Pete Zaitcev
> In a level (constant-altitude) turn at a given bank angle, does the turn rate
> (in degrees per second) vary with airspeed? An explanation would be
> appreciated.
>
Newsgroups that are more infested with cheating students who are too
lazy to move their ass to the library adopted various attitudes,
ranging from name calling or ignoring to something more funny.
One good thing to do is to post "please let us know what is your
college and course is so that we can help you better" - you would
be surprised to know that some idiots fail to that trap (well,
if they post homeworks questions, what do you expect.... :).
In comp.arch there is an agreement to post plausible but bogus
explanations, unfortunately you never know what happens when the
subject of the joke submits his work!
--Pete
Frederic Woodbridge, III
<slrn8q3a7l....@js006.noname.ru>:
>Newsgroups that are more infested with cheating students [snip]
I wouldn't go so far as to infer that questions of the sort Dr. Lowry
answered would fall under the cheating student category although I
wouldn't rule it out entirely.
It may well be a pilot trying to understand his working environs a
little better so give the guy a break :)
--
Frédéric Woodbridge, III
"A man's own tongue may cut his throat" - Sa'di
Roy Smith
> Newsgroups that are more infested with cheating students who are too
Why is it that whenever anybody asks a question here, they get pounced on with
either "you're too lazy to look it up yourself", or "your instructor must be a
total bozo if you havn't learned that yet"?
--
hsaw...@my-deja.com
it, you arrogant, paranoid freak.
If not, then I apologize for misinterpreting your reply to my message, which
I still don't understand.
If you were trolling, good job!
FWIW, I am a 2.6 hour student pilot trying to learn everything I can about
level turns, and I sincerely thank the posters who have helped me with that.
If I had the time, could I have found this answer in the library or worked
it out myself? Absolutely. (In fact I since have). Would that have led to
any human interaction/discussion on the topic, or helped anyone else out?
Nope.
The comp.arch newsgroup sounds like a real friendly, helpful community -- I
hope things never get that hostile here.
In article <slrn8q3a7l....@js006.noname.ru>,
hsaw...@my-deja.com
theory is that the virtual anonymity of the internet allows meek, nerdy,
anal-retentive types to adopt an air of superiority and condescension that,
if displayed in a face-to-face discussion, would probably get their asses
kicked. I imagine it is quite a thrill for them.
PS Thanks again for all your help Mr. Smith.
In article <roy-656B0A.0...@netnews.nyu.edu>,
Roy Smith <r...@popmail.med.nyu.edu> wrote:
> zai...@yahoo.com (Pete Zaitcev) wrote:
> > Newsgroups that are more infested with cheating students who are too
> > lazy to move their ass to the library [blah, blah, blah]
>
> Why is it that whenever anybody asks a question here, they get pounced on with
> either "you're too lazy to look it up yourself", or "your instructor must be a
> total bozo if you havn't learned that yet"?
>
> --
> Roy Smith <r...@popmail.med.nyu.edu>
> CP-ASEL-IA, CFI-ASE-IA
>
>
James Wilkinson
anal-retentive OR to adopt an air of superiority and condescension. Or to
get my ass kicked.
<hsaw...@my-deja.com> wrote in message news:8nu1ju$eod$1...@nnrp1.deja.com...
> At the risk of being accused of being a cheating psychology student, my
> theory is that the virtual anonymity of the internet allows meek, nerdy,
> anal-retentive types to adopt an air of superiority and condescension
that,
> if displayed in a face-to-face discussion, would probably get their asses
> kicked. I imagine it is quite a thrill for them.
>
> PS Thanks again for all your help Mr. Smith.
>
> In article <roy-656B0A.0...@netnews.nyu.edu>,
> Roy Smith <r...@popmail.med.nyu.edu> wrote:
> > zai...@yahoo.com (Pete Zaitcev) wrote:
> > > Newsgroups that are more infested with cheating students who are too
> > > lazy to move their ass to the library [blah, blah, blah]
> >
> > Why is it that whenever anybody asks a question here, they get pounced
on with
> > either "you're too lazy to look it up yourself", or "your instructor
must be a
> > total bozo if you havn't learned that yet"?
> >
> > --
> > Roy Smith <r...@popmail.med.nyu.edu>
> > CP-ASEL-IA, CFI-ASE-IA
> >
> >
>
>
Bob Gardner
FAA when making up test questions. On page 179 there is a nomograph of speed,
bank angle, turn radius, etc.
Bob Gardner
hsaw...@my-deja.com wrote:
> In a level (constant-altitude) turn at a given bank angle, does the turn rate
> (in degrees per second) vary with airspeed? An explanation would be
> appreciated.
>
> thanks
>
Jeffrey Osier-Mixon
> I resent that. I don't need the internet to be meek, nerdy, and
> anal-retentive OR to adopt an air of superiority and condescension. Or to
> get my ass kicked.
Hear hear! "My skin ain't green, but I'm still good and mean."
Regarding two-minute turns... I have been dividing them into 4ths and basically
doing four 30-second 90-degree turns. I find that this helps in windy
conditions, and it satisfies the requirement of turning 360 in 2 minutes. Plus,
it helps me prevent myself from over-correcting ALL the time, this way I only do
it four times during that maneuver.
R.S.
airspeed in knots. 100 knots => ~15 deg. of bank. In the instrument world,
we tend to limit bank angles to about 30 deg. This corresponds with a
standard rate turn at 200 knots. Beyond that speed, limit turns to 30 deg
bank or less, and accept less than standard rate turn.
Ray Suziedelis, CFII, MEI
Robert Frischer wrote in message
<00ho5.46973$2k1.1...@news-west.usenetserver.com>...
>A standard rate means that you do a 360 in 2 minutes. Doesn't matter what
>the airspeed is. The faster plane will have a larger turning radius to do
>it. Standard rate turns are done with the turn coordinator.
>
>--
>Robert Frischer, rfri...@iname.com
>"Tracy R Reed" <tr...@freeside.ultraviolet.org> wrote in message
>news:slrn8q348j...@freeside.ultraviolet.org...
>> Roy Smith <r...@popmail.med.nyu.edu> wrote:
>> >degree bank angle, again with coordinated rudder. Compare the size of
>the
>> >circle you make and the amount of time it takes to do 360 degrees of
>heading
>> >change with what you got before.
>>
Mark Burkley
wrote:
>
>Turn rate (in radians per second) = g(in ft/sec^2)*tangent(bank angle)/true
>air speed (in ft/sec). A radian is about 57.3 degrees.
Hey, cool formula. I've no idea how you derived it despite Roy's hint
:-) (Guess I did "first semester calculus" too long ago).
But most of the terms are constants. It might be more useful if we
converted everything to metres & seconds, multiplied up all the
constants and put bank angle on the left.
So, for a rate one turn (2pi/120) and g is 9.81 m/s^2 and TAS in nm/h
is 1852 x TAS / 3600 m/s:
bank angle = atan(tas/364)
At 100 kts, angle is 15 degrees, 200 knots, angle is 29 degrees. 300
knots is 40 degrees.
Sound about right?
--
Mark Burkley mark.b...@intel.com
Intel Integrated Access Division direct: +353 61 716026
Shannon, Ireland fax: +353 61 716006
R.S.
Ray Suziedelis, CFII, MEI
highflyer
>
> In a level (constant-altitude) turn at a given bank angle, does the turn rate
> (in degrees per second) vary with airspeed? An explanation would be
> appreciated.
>
In a word, yes.
The bank angle determines the percentage of your lift that is redirected
toward the center of the turn. This redirected lift provides a force
that gives you an accelleration to redirect your momentum vector to
allow
you to change direction. This force is directly proportional to your
mass. The central accelleration is directly proportional to the force.
However, the momentum vector you are trying to modify is proportional
to the square of your velocity. Momentum is kinetic energy which is
1/2 mass times velocity squared.
A your speed increases the momentum change you have to make for a
given number of degrees of turn increases with the square of your
speed. As a result, you need more central accelleration to change
the direction of the momentum vector at a constant rate.
You can check this out by making constant rate turns, with the needle
in the doghouse, at different speeds. You will find that your bank
angle increases as your speed increases. Airliners, flying at
relatively higher speeds, usually make less than standard rate turns
to keep the bank angles down to comfortable levels for the passengers.
--
HighFlyer
Highflight Aviation Services
highflyer
>
> Roy Smith <r...@popmail.med.nyu.edu> wrote:
> >degree bank angle, again with coordinated rudder. Compare the size of the
> >circle you make and the amount of time it takes to do 360 degrees of heading
> >change with what you got before.
>
> So, the faster you go, the longer it will take you to do a 360 degree turn.
> This implies that a "standard rate turn" is not the same in every aircraft,
> correct? I have not yet begun my instrument training yet but I have been told
> that 2 minute turns are the thing to do in IMC and my little 152 seems happy
> enough doing a 2 minute 360 degree turn. How long would a 747 take to do a 360?
>
a "standard rate turn" is a standard rate turn. The two minute turn
takes
two minutes. No matter what your speed is. However, at higher speeds
it
will take a much greater bank angle to make a two minute turn. A 747
CAN
make a "two minute" turn. However, the bank angle is enough to give a
noticeable increase in apparent "weight" for the passengers because of
the increased G loading, so they don't normally make "standard rate"
turns.
In the C150 the "standard rate" turn is a very comfortable bank angle.
Roy Smith
> Airliners, flying at relatively higher speeds, usually make less than
> standard rate turns to keep the bank angles down to comfortable levels for
> the passengers.
Is that really the reason? I thought the problem was that at high altitude
cruise, the stall margin is so small that even relatively modest load factors
could cause control problems.
Zachary Kessin
> hsaw...@my-deja.com wrote:
> >
> > In a level (constant-altitude) turn at a given bank angle, does the turn rate
> > (in degrees per second) vary with airspeed? An explanation would be
> > appreciated.
> >
>
> In a word, yes.
>
> The bank angle determines the percentage of your lift that is redirected
> toward the center of the turn. This redirected lift provides a force
> that gives you an accelleration to redirect your momentum vector to
> allow
> you to change direction. This force is directly proportional to your
> mass. The central accelleration is directly proportional to the force.
>
> However, the momentum vector you are trying to modify is proportional
> to the square of your velocity. Momentum is kinetic energy which is
> 1/2 mass times velocity squared.
Um no, Momentum is mv, and is a vector, Kenetic Energy is (mv^2)/2 and
is a scalar value. They are very different. (Infact momentum is the
first order dervitive of Kenetic Energy.)
>
> A your speed increases the momentum change you have to make for a
> given number of degrees of turn increases with the square of your
> speed. As a result, you need more central accelleration to change
> the direction of the momentum vector at a constant rate.
>
Newton's Second law, F=M*A -> F=M*DV/DT
--Zach
(one time physics geek.)
> You can check this out by making constant rate turns, with the needle
> in the doghouse, at different speeds. You will find that your bank
> angle increases as your speed increases. Airliners, flying at
> relatively higher speeds, usually make less than standard rate turns
> to keep the bank angles down to comfortable levels for the passengers.
>