PDA

View Full Version : AoB / rate of turn question


KillKenny
13th Feb 2002, 17:21
Angle of Bank for a Rate 1 turn = [TAS/10]+7, right?

Is AoB for a Rate 2 = [2xTAS/10]+7???

Paul Hickley
13th Feb 2002, 19:49
That formula is actually a bit of an approximation.

If you want a full reply to this question from a JAA ATPL professional ground school instructor, I suggest that you post it on <a href="http://www.oxfordaviation.net/forum/forum.asp?FORUM_ID=15" target="_blank">@SK OXFORD FORUMS</a>.

All the best,

Paul Hickley

Keith.Williams.
14th Feb 2002, 00:33
KillKenny,

The short answer is no, and it isn't just because the equation is an approximation!

The problem is that rate of turn is related to TAS squared/Tangent of AOB, neither of which are linear functions. So you cannot simply extrapolate the original equation in a linear manner. The fact is that the equation you quote is simply a trick which happens to work for a rate one turn. But it works only for a rate one turn. One could however be devised for any rate of turn we might choose.

To demonstrate my point let's look at a rate 8.825 turn (whatever that is) Let's use 400 Kts as an arbitrary TAS.

We will then get:

AOB ((8.825 x 400)/10)+7 which = 360 degrees.

But this means that you can do a rate 8.825 turn at 400 kts with the wings level. Hummmmmm?

Worse still, a rate 8 turn turn requires 327 degrees AOB. This means that to do a rate 8 turn to the right we must bank 33 degrees to the left??? Even bigger Hummmmmmmmmmmm? (even I don't believe that).

Your e-mail address is not listed in your profile. If you send it to me I will send you a scan of a nomogram that enable you to wark out AOB for any combination of TAS and ROT up to rate 4.

[ 13 February 2002: Message edited by: Keith Williams. ]</p>

john_tullamarine
14th Feb 2002, 05:33
The basic turn performance equation is. .

radius = V^2 / (g * tan(bank angle))

which can be massaged (by getting rid of the constants) into

radius = KTAS^2 / (68637 * tan(bank angle))

for radius in nm, bank angle in degrees, and speed in kt.

If some more work is done on the first equation, we can arrive at

rate of turn = (g * tan(bank angle)) / V

which can be massaged into

rate of turn = 1092 * tan(bank angle) / KTAS

for rate of turn in degrees/second, bank angle in degrees, and speed in kt.

Both of these equations you can rearrange to suit your requirements and they suit normal calculator sums just fine. If you are really, really keen and into mental self abuse you can even use your Jepp or Dalton nav computer to confuse you with the answer ...

You may find equations elsewhere which have the multiplying factors differing to the above by a very small amount ... this just reflects the specific constant values used and calculator round off errors.

Alternatively, you may find equations in a different form, eg it is quite common to find the rate of turn equation expressed in terms of load factor, which is not a great deal of use for a civil pilot .. in any case, that sort of presentation reduces to the above equation with just a modicum of trigonometric massaging.

The basic derivation of these equations is quite straightforward and, should you have a boring need to know it, is shown in just about every basic undergrad aerodynamics or flight mechanics textbook ... one of which should be available in your local library.

Likewise, if you want a nice, easy to use graph, I can email you one I drew up for a local IFR training school some years ago.

The workaround described in the first post is a bit rough and ready, giving reasonable answers only for the normal sort of letdown procedural speeds ....

. .For practical considerations in civil IFR operations, I really can't see much use in worrying too much about this stuff .. except for the very occasional base turn letdown where the procedure designer faces a problem (or makes a mistake) which results in too tight a turn radius, 20 to 30 degrees of bank (depending on the aircraft's speed) works fine provided appropriate allowance is made for a sensible lead procedure to acquire the next track.

[ 14 February 2002: Message edited by: john_tullamarine ]</p>

BEagle
14th Feb 2002, 12:06
Or to make drawing maps at low level easy, use the hole in the end of a standard RAF navigation ruler to draw the turn and then work out what bank angle you need at different speeds to get round the turn you've already drawn!!

Keith.Williams.
15th Feb 2002, 00:35
JT,

I cannot argue with any of your maths, and all this stuff certainly has little relevance to a practicing ATPL. But to an ATPL student it is essential stuff. If KK is studying for his technical exams, or has any thoughts of doing so in the future, researching this kind of stuff can be nothing but helpful.

Similarly, the nomogram I referred to will be of little value in the cockpit, but if he wants to try to devise his own shortcut equations, or modify any he gets from reference books, it will enable him to test them without lots of button pushing.

[ 14 February 2002: Message edited by: Keith Williams. ]</p>

john_tullamarine
15th Feb 2002, 04:10
Keith,

I concur that exam workups are a little removed from the real world but that has always been the way of the game.

Certainly the basic relations are essential for the initial IFR trainee .. makes for an easier transition to getting on top of letdown manoeuvring.

The use of the various graphical representations which one sees around the place makes it all far easier for the student .. certainly number crunching is not everybody's cup of tea.

Chuck Ellsworth
15th Feb 2002, 07:19
Hey you guys:

I am working at a bit of a disadvantage here because I wrote my ALTP back in the days when the answers were hand written, not multiple choice. And slowly the need to remember all these equations have faded from memory as I have gotten complacent and just fly the airplane without to much thought about the math part.

Anyhow there is one equation I have forgotten, maybe someone here can refresh my memory.

If friction produces heat, how long will it take a dog to hump a pail of water to a boil? <img src="smile.gif" border="0"> <img src="smile.gif" border="0"> <img src="smile.gif" border="0">

..............

:) The hardest thing about flying is knowing when to say no :)

john_tullamarine
15th Feb 2002, 12:44
Cat Driver,

I really was hoping that you would ask that question ... although the answer involves some higher mathematics ...and would bore the pants off most of our colleagues ..

Reminds me of two questions posed in my undergrad years in similar vein but which took a lengthy solution to get a tick ...

(a) my fluid mech lecturer posed the following tutorial question along the lines of .. a wartime submarine is disabled on the bottom (lots of data about the sub and depths etc ..) .. how do we get the guys out and to the surface without any third party assistance. The typical answer rambled on via several reams of paper, numerous sums and suchlike ... and, at the end, I penned an impertinent comment to the effect that they would be dead during the ascent anyway due to the depth and other factors ... it was a little embarassing when I got the marked paper back and found his comment "I'm not".

(b) the same guy, in the final exam posed this question. (Background, the old PNR building at SU had a lovely old spiral staircase going up however many floors) .... should the tea lady, having made the tea on the ground floor and put in the sugar ... stir it ... or just walk up the stairs to the top floor .... that answer went on and on also ....

However ... I reminisce and digress ... as usual ..

KillKenny
15th Feb 2002, 14:02
Thanks for all your responses guys. Original post was made in moment of idiocy.

john_tullamarine
15th Feb 2002, 14:14
Do be relaxed KK ... we all have our moments ...

Keith.Williams.
16th Feb 2002, 16:44
KK,

Your question is not a sign of lunacy; it's just an attempt to explore the bounds of the equation.

To illustrate my point that it is only good for rate 1 turns, I have run the shortcut equation and a more accurate one through a spreadsheet. The results were as follows:

Comparing the results for Rate 1 turns at various speeds

TAS TAN AOB = TAS x ROT AOB = (TAS/10)+7 . .(Kts) g . . Error . .180 Kts, 26 actual AOB, 25 prediected, -1 error. .200 Kts, 28.4 actual AOB, 27 predicted, -1.4 error. .250 Kts, 34.1 actiual AOB,32 predicted, -2.1 error. .300 Kts, 39 actual AOB, 37 predicted, -2 error . .350 Kts, 43.4 actual AOB, 42 predicted, -1.4 error. .400 Kts, 47.2 actual AOB, 47 [redicted, -0.2 error. .450 Kts, 50.6 actual AOB, 52 predicted, +1.4 error. .500 Kts, 53.5 actual AOB, 57 predicted, +3.5 error. .550 Kts, 56.1 actual AOB, 62 predicted, 5.9 error. .600 Kts, 58.4 actual AOB, 67 predicted, 8.6 error

This shows that the shortcut equation is reasonably accurate throughout a fairly wide speed range.

. .Comparing results for various rates of turn at 250 Kts TAS.

Turn rate TANAOB = TAS x ROT AOB=((N xTAS)/10)+7 (N) g . . . .Rate 0.5, 18.8 actual,19.5 predicted, 0.7 error. .Rate 1.0, 34.1 actual, 32 predicted, -2.1 error. .Rate 1.5, 44.6 actual, 44.5 predicted, -0.1 error. .Rate 2.0, 53.7 actual, 57 predicted, 3.3 error. .Rate 2.5, 59.6 actual, 69.5 predicted, 9.9 error. .Rate 3.0, 63.9 actual, 82 predicted, 18.1 error. .Rate 3.5, 67.2 actual, 94.5 predicted, 27.3 error. .Rate 4.0, 69.8 actual, 107 predicted, 37.2 error

The shortcut equation is reasonably accurate only within a very narrow band (up to rate 2 turn), and then quickly diverges such that its results are nonsensical (more than 90 degrees AOB) at rate 3.5 and above.

. .CAT DRIVER,

Your problem is a bit more complicated.

First, we need to know the rate at which heat is put into the system. This is unlikely to be a linear function, because it will be proportional to the hump rate. So let's use the derivative dh/dHR.

However, hump rate is not likely to be constant either. For very young dogs, it will increase rapidly, whereas for very old dogs it will quickly subside. To account for this we require another derivative, dHR/D ageofdedog.

Next we need to consider the rate at which the system cools due to the escape of heat. Once again, it will not be linear so we need to use the derivative dc/dt.

Therefore, we now have an equation something like:

Temperature = dh/DHR . dHR/D agededog . dc/dt

Then of course we need to consider the quantity of water in the system. The problem here is that however much we might try to maintain constant experimental conditions, it is inevitable that someone will come along and take out some of the water.

This phenomenon is admirably illustrated by JT's conundrum about the sunken submarine. For some desperate people extracting water is the only way to get it up!!

Is this perhaps the problem you really wanted to share with us?

[ 16 February 2002: Message edited by: Keith Williams. ]</p>

allthenines
16th Feb 2002, 21:28
Any idea why engineers use nitrogen in aircraft tyres? <img src="rolleyes.gif" border="0">

allthenines
16th Feb 2002, 21:32
Sorry guys I meant to pose this question as a new topic. Disregard. <img src="eek.gif" border="0">

john_tullamarine
17th Feb 2002, 04:39
Keith,

I think that I should like to sit in on one of your lectures one day .... I suspect that you would be near as entertaining as Julius ("Why is it so") Sumner Miller was wont to be .. and he was a most entertaining fellow ...