What is the formula for this?

[Gary Warner]

Given a known constant bank angle and a known constant airspeed, what's the formula for calculating the diameter (or radius) of a level 360 deg. turn?

(formula to be used in the next topic I post)

Gary
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[Sail 'n Soar]

Given a known constant bank angle and a known constant airspeed, what's the formula for calculating the diameter (or radius) of a level 360 deg. turn?

(formula to be used in the next topic I post)

Gary
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R = V^2*tan(theta)/g

where:
R = radius of a level turn
V = velocity in fps
theta = bank angle
g = acceleeration of gravity

If V in mph and R is in feet, the approximate equation is:

R = .0668*tan(theta)*V^2

[erik1938]

Sail n soar. Did you derive that formula or obtain it from a book? I guess I am curious at to how this is derived, because each plane may have different flight characteristics.

If the bank angle is constant(ailerons are fixed) and the elevator is used to tighten the turn, then this formula wouldn't apply. Maybe I am not thinking of the correct situation.

[Sparky Paul]

Sail n soar. Did you derive that formula or obtain it from a book? I guess I am curious at to how this is derived, because each plane may have different flight characteristics.

If the bank angle is constant(ailerons are fixed) and the elevator is used to tighten the turn, then this formula wouldn't apply. Maybe I am not thinking of the correct situation.
.
???
Why wouldn't it?
The condition of "level" sets the other pararmeters. Speed and g and bank angle depend on "level".

[erik1938]

Ok, I plotted this formula in excell and as you can see from the formula it depends entirely on the bank angle.

At 90 a degree bank angle, the radius would be approx 4.35E17 ft, and at 45 degrees the radius would be 26.6 ft. The values I used were:

V = 29.3 ft/s
g = 32.2 ft/s^2 (constant acceleration of gravity)

So from this I don't see how this formula would work. If you think it is correct please explain the math/physics behind its derivation.
(please note I dont mean to sound like an a-hole, I just was curious as to whether or not this is correct.)

[vintage1]

It is correct. Assuming the lift is perpendicular to the wing, then the total lift is split into two vectors. One keeps the model in a level turn, the othert is the G force being applied by the turn. What you are missing is the fact that if you apply up elevator to tighten the turn, the model will climb unless its banked over a bit more...

Think motortcycle. To go round a turn faster you have to lean over more, to go round a turn tighter you have to lean over more...

[Gary Warner]

Thanks for the formula. Now I'm off to try and smoke my calculator batteries :).

Gary
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[flieslikeabeagle]

Sail 'n Soar posted the formula:

R = V^2*tan(theta)/g

where:
R = radius of a level turn
V = velocity in fps
theta = bank angle
g = acceleeration of gravity


This formula uses the *complement* to the bank angle, i.e. in the formula a level wing corresponds to theta = 90 deg rather than theta = 0. (Easy way to check: stick in theta = 0, this predicts R =0 since tan(0) = 0. In other words, the formula says the turn is infinitely tight at theta = 0. This also matches erik1938's Excel result of a (nearly) infinitely large radius turn at theta = 90 deg ).

Using the conventional definition of bank angle (i.e. a level wing corresponds to theta = 0 rather than theta = 90 deg), the formula changes to

R = V^2/g * cot(theta)

i.e., replace tan (theta) in the original equation with cot (theta), which is the reciprocal of tan(theta).

-Flieslikeabeagle

[FlyByMike]

My father (retired USAAF & USAF pilot) often talked of a "standard turn". IIRC he was saying something about banking a fixed amount (marked on his instruments? or having his flying wires parallel to the ground?) would result in a turn that would take exactly 60 seconds.
Can anyone help in the specifics and meaning or importance of this?
And does it add to this thread ?
-Mike

[flieslikeabeagle]

FlyByMike, I have no knowledge whatsoever of a "standard turn", but I played with the formula for bank angle and the basic formula connecting distance, time, and speed, and I found out that the condition:

V cot(theta) = constant

...will result in any turn through a specific number of degrees taking the same time (i.e. a 90 degree turn will always take the same time, irrespective of the speed V), and simultaneously keep the plane at the proper bank angle. The constant determines how many seconds long that time is.

In other words, this formula tells you at what angle to bank for any specific speed your plane is moving, and the result will be that any 90-deg (or 20-deg, or 360-deg, etc) turn will take the same time.

The formula can be re-arranged to look like this:

theta = atan(V/constant)

meaning that for any specific speed V, there is one specific bank angle theta that results in a constant-time turn.

Certainly you could in principle put marks on the air-speed indicator telling you at what angle to bank for any specific airspeed, matching the formula above. Whether this was ever actually done, I have no idea. Hopefully others on this forum will...

-Flieslikeabeagle

[Sparky Paul]

There's a "standard rate turn"... one needle width on the turn and bank indicator. ISTR it's 2 minutes around.

[NigelD]

When the needle at the top is aligned with the marks to the left or right, The aircraft is turning at 3deg/second. I think aircraft use these turns when in a holding pattern. There is no bank angle indicator there, the thing in the middle is a balance ball, just like a spirit level. the pilot aims to keep the ball in the centre by rudder coordination.

so i guess this instrument just takes an airspeed reading and bank angle reading from other instruments and gives this turn rate??
Or does it take climbing or decending into account and give a true reading?

ive just realised that the balance ball must be in the centre for this to work i.e. no slip.
A plane can be flown in a straight line horizontally with any bank angle... eg knife edge!

[HarryC]

so i guess this instrument just takes an airspeed reading and bank angle reading from other instruments and gives this turn rate??

No, it's simply gyrosocopic, it shows an angular rate of change. It is therefore independent of airspeed and bank angle. It swings around while taxying with the aircraft level, it just shows how fast the nose is changing heading left or right. Whatever speed you are doing, you simply turn less or more tightly as required to keep the needle on the mark. This gives aircraft a standard rate of change of heading. The radius of the turn will then depend upon the airspeed, at slow speed it will be a small turn with less angle of bank and at high speed it will be a large radius but at a steeper angle of bank and commensurately higher G load.

H