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 Post subject: Rate of Turn Equation for Aircraft.
PostPosted: Wed, 16 May 2012 19:59:59 UTC 
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I have this equation: Tanx = V^2/gr or r = V^2/gtanx

How do I derive the equation of Rate of Turn using the above equation?

The equation result would be: ROT = gtan(x)/V(feet per second) The equation calculates ROT in Radians/sec.

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 Post subject: Re: Rate of Turn Equation for Aircraft.
PostPosted: Wed, 16 May 2012 23:12:10 UTC 
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ShawnGroskreutz wrote:
I have this equation: Tanx = V^2/gr or r = V^2/gtanx

How do I derive the equation of Rate of Turn using the above equation?

The equation result would be: ROT = gtan(x)/V(feet per second) The equation calculates ROT in Radians/sec.


Without knowing what any of those variables are, who knows?

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 Post subject: Re: Rate of Turn Equation for Aircraft.
PostPosted: Thu, 17 May 2012 03:00:56 UTC 
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r is radius in feet
g gravity 32.2m/s

V Feet/Sec

Tan(x) Tan of the Bank Angle

My best attempt is:

D=Vt

t = D/V and C = 2pi(r)

t = 2pi(r)/V and r=V^2/gtan(x)

Thus: t = 2pi[(V^2/gtan(x))]/V

And: t = 2(pi)V/gtan(x)

However, I think I may be missing some important steps. The ROT is not taken into account and I am not sure how to make an equation for ROT to insert into the main equation.

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 Post subject: Re: Rate of Turn Equation for Aircraft.
PostPosted: Thu, 17 May 2012 08:04:40 UTC 
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ShawnGroskreutz wrote:
r is radius in feet
g gravity 32.2m/s

V Feet/Sec

Tan(x) Tan of the Bank Angle

My best attempt is:

D=Vt

t = D/V and C = 2pi(r)

t = 2pi(r)/V and r=V^2/gtan(x)

Thus: t = 2pi[(V^2/gtan(x))]/V

And: t = 2(pi)V/gtan(x)

However, I think I may be missing some important steps. The ROT is not taken into account and I am not sure how to make an equation for ROT to insert into the main equation.


Presumably your v represents the equivalent linear velocity at radius r where you have a purely rotational motion, in which case use v=r\omega to get \omega=\dfrac{v}{r}=\dfrac{g\tan x}{v}.

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject: Re: Rate of Turn Equation for Aircraft.
PostPosted: Thu, 17 May 2012 13:43:18 UTC 
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Ok, that was a lot easier. Did not think in that direction.

But, was I on the right track with my work, what did I need to do finish the work?

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