Aces High Bulletin Board
General Forums => Aces High General Discussion => Topic started by: clerick on August 04, 2007, 04:55:49 PM
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When i was unpacking some books i ran across an old one that came with Falcon Gold WAY back in the 90's. It's called "Art of the Kill", interesting book but in it the author mentions a formula for turn rate
TR = K G/V
G = the g's the aircraft is pulling in the turn
V = is velocity
but K is only listed as a constant. I would assume that this constant varies from plane to plane. Could any of you explain what K represents and how it is derived?
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Originally posted by clerick
When i was unpacking some books i ran across an old one that came with Falcon Gold WAY back in the 90's. It's called "Art of the Kill", interesting book but in it the author mentions a formula for turn rate
TR = K G/V
G = the g's the aircraft is pulling in the turn
V = is velocity
but K is only listed as a constant. I would assume that this constant varies from plane to plane. Could any of you explain what K represents and how it is derived?
*see widewing!:aok
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LOL.... now that you mention it i was kind of hoping he would drop by and enlighten me. :D
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Mosq !
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Originally posted by Angry Samoan
Mosq !
riiiiiiiiiight..... move along now, GIT!
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Originally posted by clerick
When i was unpacking some books i ran across an old one that came with Falcon Gold WAY back in the 90's. It's called "Art of the Kill", interesting book but in it the author mentions a formula for turn rate
TR = K G/V
G = the g's the aircraft is pulling in the turn
V = is velocity
but K is only listed as a constant. I would assume that this constant varies from plane to plane. Could any of you explain what K represents and how it is derived?
Could be "Drag due to lift"--aka induced drag. A formula for K is
K = 1 / pi*e*A, where pi equals 3.14, e is Oswald's Span Efficiency Factor, and A is aspect ratio. Might want to simply substitute a value between .75 and .85 for e. Aspect ratio should be the wingspan^2 / wing area in ft^2...
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TR = K * g / V
Hmmm... that's a funny equation. Are you sure you have transposed it here correctly?
The only turn rate equation that is remotely similar that I'm aware of is:
TR = tan(b) * g / V
This is as close to your's as I can get by replacing K with tan(b). In this equation g isn't g-load but gravitational acceleration (a constant). (b) is bank angle.
tan(b) is usually substituted as a function of load-factor n (g-load of the aircraft). This form of the turn rate equation is:
(http://brauncomustangs.org/images/eq1-2.jpg)
Tango, XO
412th FS Braunco Mustangs
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Originally posted by dtango
TR = K * g / V
Hmmm... that's a funny equation. Are you sure you have transposed it here correctly?
The only turn rate equation that is remotely similar that I'm aware of is:
TR = tan(b) * g / V
This is as close to your's as I can get by replacing K with tan(b). In this equation g isn't g-load but gravitational acceleration (a constant). (b) is bank angle.
tan(b) is usually substituted as a function of load-factor n (g-load of the aircraft). This form of the turn rate equation is:
(http://brauncomustangs.org/images/eq1-2.jpg)
Tango, XO
412th FS Braunco Mustangs
That is the formula I'm aware of as well.
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So like, 1+1 is still 3 right?
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Clerick, give a look at Brooke's page about Math behind turning:
http://www.electraforge.com/brooke/flightsims/aces_high/stallSpeedMath/turningMath.html
I think you'll find there all you want to know. :)
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The maths behind displacing inertia into a different direction with least loss is...Look at how a noob tries to do it, then relate how you know how to do it, and call it intuitive experience that defines mathematics...(when you getting shot at often makes mathematical principles quite clear aand most times rather accidentally...though one has to reliaze that Aces High is a computor model trying to relate to the real thing)
So maybe its best to experiment with the Aces High model of all these forces and maybe realize, it quite likely does not represent the 'real deal'
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Who cares? In the middle of a fight it doesn't matter if I've got a mathematical chance of winning or not, I'm still going to die.
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It doesn't seem to me that clerick asked anything about AH flight model or said he wanted to calculate his turn rate in the middle of a fight. He just plainly asked for a real life formula.
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In light of my scientific analysis of the above conversation, I have come to one obvious conclusion.
1. I am a COMPLETE idiot in regards to math.
Although there are some who would say I'm just a complete idiot.
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I'm an engineering student and as such i tend to be drawn to formulas. I asked the question merely out of curiosity.
I did double check the formula and I presented it exactly as written in the book. What i don't know for sure is if "G" is indicated g's in the turn or radial g. I would assume that it is radial g.
Looking at Gianlupo's website i think I've found my "K".
omega = (g / v) * sqrt[(L/W)^2 - 1]
it would appear that sqrt[(L/W)^2 - 1] = K. Though the author uses g instead of G. I can's see where he defines g. Classical physics would say that it is the force due to gravity, but looking at the formula the website offers i will assume that it is sloppy math since his formula appears to disregard the g's the aircraft is pulling.
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Originally posted by clerick
When i was unpacking some books i ran across an old one that came with Falcon Gold WAY back in the 90's. It's called "Art of the Kill", interesting book but in it the author mentions a formula for turn rate
TR = K G/V
G = the g's the aircraft is pulling in the turn
V = is velocity
but K is only listed as a constant. I would assume that this constant varies from plane to plane. Could any of you explain what K represents and how it is derived?
Hi clerick
That formula is perfectly correct. Pete Bonanni wouldn't get that sort of thing wrong. He is simply using a slightly different definition of the terms than those normally used in order to keep the expression simple in order to focus on the concepts. The value you are looking for is K = 1257 when the speed is in miles per hour.
So if you are turning at 4g and 200mph your turn rate will be:
1257*4/200 = 25 degrees per second (dps)
The more complex formula others have posted would yield 24.3dps. The difference is due to the fact that Bonanni is using Radial G whereas the formula others have posted are actually using the Body axis normal load factor, and not the radial G.
If I use typical spread sheet math symbols and use n for the body axis normal load factor, the radial G can be found by using Pythagoras's theorem as G = SQRT(n^2 - 1). that's where that term under the square root comes from in those other equations.
Hope that helps.
Badboy
Edit, forgot to mention where the 1257 came from:
It's just gravity times a conversion factor to get miles per hour to ft/sec and another conversion factor to get radians per second to degrees per second. Using spread sheet math again it looks like this:
=32.185/(5280/(60*60))*(180/PI()) = 1257.3168
You should be able to see what each of those numbers represents.
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Ah, thanks for clarifying Badboy. Radial G vs. normal load factor.
Tango, XO
412th FS Braunco Mustangs
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Badboy, question: what do we read on the g-meter? The radial G or the body normal axis load factor?
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Originally posted by Gianlupo
Badboy, question: what do we read on the g-meter? The radial G or the body normal axis load factor?
The g-meter will read the body axis normal load factor, because that’s the way it is orientated, which will be slightly less than the load factor due to lift (L/W) which in turn will be slightly more than the radial G if the turn is carried out at constant altitude.
Badboy
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Greetings,
My brain hurts.......where's the asprin.
Regards,
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thanks for the help
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Originally posted by Badboy
The g-meter will read the body axis normal load factor, because that’s the way it is orientated, which will be slightly less than the load factor due to lift (L/W) which in turn will be slightly more than the radial G if the turn is carried out at constant altitude.
Badboy
Ok, thank you, but... another question arises. I thought the body axis normal factor was L/W, clearly it's not. So, what is the body axis normal factor? And how is it mathematically defined? Thanks for the patience, Badboy! ;)
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I wanted to do AeroEng at one point.
I can only say, in hindsight, that it's a darn good thing for the air-going public that I didn't.
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Originally posted by Gianlupo
Ok, thank you, but... another question arises. I thought the body axis normal factor was L/W, clearly it's not. So, what is the body axis normal factor? And how is it mathematically defined? Thanks for the patience, Badboy! ;)
They are almost the same, but not quite, because the lift acts normal to the free airstream, and the aircraft axis is tilted back relative to that due to angle of attack. Because the AoA is normally small, the difference is generally less than about 4%. So the G force acting normal to the aircraft's longitudinal body axis, is (L/W)*Cos(Alpha). Here is a diagram:
(http://www.badz.pwp.blueyonder.co.uk/images/Turndiag.jpg)
EDIT:
Sorry about the F-16s, I originally drew the diagram for a discussion about Jets... Also just to correct myself, what Bonanni has done in his simplified equation is to assume that the AoA is negligible and can be ignored, and that the aircraft is banked at 90 degrees, so that the one vertical G isn't needed. You get, a slightly better turn rate by doing that in practice, at the expense of loss in altitude.
Hope that helps.
Badboy
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hmmm interesting thread. Shame Gian's avatar is distracting my line of thought.
I'd hit that !
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Got it, Badboy, thank you again.
Bruv, the purpose of my avatar is just to show my actual location on an innovative kind of geographic map. :D
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Maybe i can help explain the difference between "axis body normal" and "radial" g in a different way.
In the fantastic multi media diagram i "made" the aircraft is pulling 5 "axis body normal" g's, the numbers i put on the diagram show the radial g's.
(http://i193.photobucket.com/albums/z39/clerick44/loop.gif)
On the entrance and exit the ABN g's are 5 but the Gr is 4. So in affect the pilot feels the full 5 g's but the aircraft is only pulling 4 "effective" or "radial" g's because gravity is pulling down the ABN.
When perfectly vertical the pilot still feels 5 g's and now the aircraft is pulling 5 effective g's because gravity is pulling perpendicular to the ABN.
At the top the pilot is still feeling 5 g's but now the aircraft is pulling 6 radial g's because gravity is pulling up along the ABN.
As i understand it, and i am open to correction, the radial g's are what actually turn the aircraft, the ABN g's are what affect the pilot.