Simple way to calculate bank angle with rate 1 turn?
 

Does anyone have a smart way of calculating bank angel in a rate one turn with only the airspeed as a known factor?

Kind Regards
Tim

TAS/10 + 7 is pretty accurate, up to 25 degrees bank (normally the limit).

BizJetJock's reply is spot on, and also implied within is a speed limit of 180 Kt for it's practical use.

Also of worthy note is that above approximately 180 kt the 45° procedure turn 'breaks down', and above that speed the 80° procedure turn is the only suitable alternative. :ok:

Regards,

Old Smokey

Caution, the following contains physics and will upset many pilots.



To calculate angle of bank for a rate one turn,

angle of bank = arctan( ( 2 x pi x TAS) / ( 9.81m/s^2 x 2 min) )

Be careful with your units (1 knot = 0.51444m/s).

Use TAS/10 + 7 and compare the two equations in Excel, you'll find less than 5% relative error between 120 to 180 kts and 340 to 590 kts (I did increments of 10 kts). Less than 10% relative error between 110 and 550 kts.

In the end, I'd say this is one of the better approximations we use.

Matthew.

I know another Matthew who talks like you do!

He's more of a Mass & Balance kind of guy though.....you two should hook up!

Matthew,

TAS/8 + 3 gives better results from 0kt up to 330kt which covers the usable speed range : 60kt, 12.4° (ultra lights) - 220kt, 31.1°.
However, dividing by 8 is harder than dividing by 10.

TAS/8 + 3 actually gives an error below 3.1% in the range 90kt - 300 kt.

Matthew. I assume by 9.81m/s you mean g. (which is of course variable). And by ^ you mean square root ( I know you did). So don't be a smart ass.

Re-entry,

in Matthew's equation, 9.81 m/s² just means g, the gravity acceleration,
considered as a constant value (here it does NOT mean the load factor, variable).

The equation is derived from expressing the centripetal acceleration in 2 different ways:
1) a = vω
centripetal acceleration = true airspeed * angular speed (in rad/s)
For rate one, angular speed is : 2π /120s

2) a/g = tan(γ)
centripetal acceleration / gravity acceleration = tangent ( bank angle )

Oh god ,why do i bother. I didn't question the mechanics. I made the simple point that g, i.e. the acceleration due to gravity, varies . It varies depending on position on the earth's surface, position of the moon, position of the sun relative to the earth. And all things being equal, if you stand on your head all your life, your head will be older than your feet when you die.
Have a nice life.

The variation of g on the planet is negligible compared to the other approximations in the formula.:8


Edited ;)

chksix. Indeed. btw it's 'negligible'. Nadal for the US Open.

I am all for accuracy but I would suggest on the average light aicraft attitude indicator that you would be pushed to get a better accuracy of plus/minus one degree on a desired bank angle - and we have not mentioned turning and acceleration errors yet! (Please no theory on this one though!).

Different of course for some EFIS displays and aircraft which use inertial systems to drive attitude informtion!

Many moons ago when I did my basic training at the College of Air Training one of the first things we were supposed to do on full panel was to "calibrate" the turn needle - ie do an accurate rate one turn to check the turn needle was giving the correct indication. If not make an adjustment etc so that when you did limited panel you knew you were getting an accurate rate one turn! Not sure how this would work out on the turn coordinator which is slightly different in principle of operation.

Hey Firefly. Rafael Nadal will win the US Open OK.
So what was your point.

You guys should get down the pub a little more often. :)

Luc, you're absolutely right. However, TAS/10 is easier to work out than TAS/8 and since an approximation is all that's desired, I'd stick with the easier math.

Re-entry. I'm not being a smart-ass and you're not being smart. Since g is an acceleration, the SI units are meters per second squared. The ^ is often used to indicate "raised to the power" so ^2 is squared, not square root. As far as the variation of g with location on the planet, you are correct. I'll leave it to you to calculate how this pertains to the AOB formula.

My apologies Matthew. Puff n call is of course right. I will back under my rock now.
Happy landings.

TAS? You do know about the pitot static systems of most ultralights? :}