Rate and radius of turn formula

Question

From this question and answer here there is a formula for both rate of turn and radius of turn. The formulas input are knots and degrees.

I have no doubt the constants in the formula (11.26 and 1,091) are conversions from SI to knots and degrees but can't seem to figure it out today.

Would someone walk-through how those constants were derived?

Thanks

Answer 1

The "real" formulas are

$$R=\frac{V^2}{g \tan \theta}$$

and

$$\omega = \frac{V}{R} = \frac{g \tan \theta}{V}$$

With $R$ in feet and $V$ in knots you get

$$g^*_R = \frac{g \cdot 1\,\mathrm{ft}}{1\,\mathrm{kt}^2} = \frac{9.81\,\mathrm{m/s^2} \cdot 0.3048\,\mathrm{m}}{(0.514\,\mathrm{m/s})^2} = 11.29$$

and with $\omega$ in degrees

$$g^*_{\omega} = \frac{g }{1\,\mathrm{kt} \cdot 1\,\mathrm{°/s}} = \frac{9.81\,\mathrm{m/s^2}}{0.514\,\mathrm{m/s} \cdot 0.01745\,\mathrm{rad/s}} = 1093$$