How to end up with Newtons in the calculation of the lift force

Question

In another StackExchange Physics question, @Rego had found the following equation to calculate the lift force generated by a simple rectangular blade.

$$F = \omega^2 L^2 l\rho\sin^2\phi$$ in which $\omega$ is the angular velocity, $L$ is the lenght of the helix, $l$ is the width of the helix, $\rho$ is the air density at normal conditions, and $\phi$ is the angular deviation of the helix related to the rotating axis. So a 4-helix propeller would lead to $F=4\omega^2 L^2 l\rho\sin^2\phi$ and so on."*

(source: Calculation for force generated by a rotating rectangular blade)

So when substituting the variables with the real values, he obtains this.

$$F=4\times(2\pi\times13000/60)^2\times(5\times10^{-2})^2\times(10^{-2})\times1.293\times(\sin1^0)^2$$$$\therefore F=0.072995N$$

However, when calculating this, I get $239.631...$. What steps have been taken between the result of this calculus and the Newton value he ended up with, and why ?

Answer 1

I have retyped your equation into Wolfram-Alpha and got the correct result.

WolframAlpha - Correct Answer

So what you wrote in your post is correct. You probably made a mistake when typing it into your calculator. It's possible you used radians instead of degrees in the final term.

I have also noticed that the units don't add up. Newtons is $ \frac{kg*m}{s^2}$ and you got $\frac{kg}{s}$

Answer 2

$$ 4\times(2\pi\times13000/60)^2\times(5\times10^{-2})^2\times(10^{-2})\times1.293 = 239.631 ... $$

You forgot to multiply by $\text{sin} (1°)$, which decreases your answer by a factor of approximately $0.00030462$.

Answer 3

If you have the correct mathematical formula, and you enter ALL quantity values in correct SI units, you will automatically end up with the correct (Newtons) units, if force is the result of the formula. It can't be anything else.

Answer 4

It seems that the width of the blade is missing. If we compare it with the 'traditional formula' for drag generated by an airfoil (assumed with $\sin \phi=1$), we have:

$$F=C_D \cdot 0.5 \rho V^2 S$$

see Wikipedia Drag Coefficient. Where $C_D$ denotes the airfoils capacity to generate drag. This value does not show up in the formala given for the place, so I'm guessing that the $C_D$ and 0.5 are ignored, or equal to 1. This leaves:

$$F= V^2 S \rho$$

If we compare this with the formula given, with $\sin \phi=1$:

$$F = \omega^2 L^2 l \rho$$

We can recognize the speed $V$ as $\omega L$ to obtain

$$F = V^2 l \rho$$ This is very alike to the simplified formula given for by @Rego

The only difference is that we have the Surface $S$, whereas Rego has the length of the foil $l$. Therefore, in Rego's formula, the width of the foil is missing. If this value were to be added, it would match the 'traditional lift formula', and will solve the units issue as well. Without width, the units are $\frac{kg}{s^2}$ (as opposed to the $\frac{kg}{s}$ presented by @chlkbumper.) With a width added, the units are $\frac{kg\cdot m}{s^2}$ or $N$

That said, I think it is a rather crude method, a large part of the calculations concerning drag consists out of finding the correct value of $C_D$. Furthermore Wikipedia gives a value of $C_D=1.28$ for a flat plate perpendicular to the flow.