How to reach this formula for the lift force of a rectangular blade There is a question made by @rego some years ago in which he uses a formula to calculate the lift force for a nylon rectangular blade. I have been researching on this but I could not find anything about this but I want to use it for a a physics essay homework. Can someone tell me how did he reached it? Or any link related to this formula. 
Here is the link of the orinigal question: Calculation for force generated by a rotating rectangular blade
Thank you. I am new here so I hope someone can help me please.
 A: DISCLAIMER - what follows is the simplified "analysis" that leads to the expression you were asking about. This is NOT in fact the correct way to treat this problem, as was pointed out in the comments.

When a blade with area $A$ moves through the air at a certain velocity $v$ and at an angle $\theta$, it "cuts through" an volume of air given by $V = A\sin\theta v$ every second - the projection of the area of the blade multiplied by the distance it moves in unit time.
This body of air is deflected downwards, and the velocity it attains is $v_d = v\sin\theta$. Now we can compute the mass of the body of air from the density and volume: $m = \rho V = \rho A \sin\theta v$. To accelerate a certain mass per unit time downwards, we need a force $F\Delta t = m\Delta v$. It follows that
$$\begin{align}F &= m\Delta v \\
&= \left(\rho A \sin\theta v\right)\left( v\sin\theta\right)\\
&=\rho A v^2 \sin^2\theta\end{align}$$
Now all you have to do is convert the linear velocity to a rotating velocity, noting that not every point on the blade will travel at the same speed when it's going around in a circle.
Note - lots of simplifying assumptions went into the above. The real physics of a blade moving through air has to take account of the fact that it's not only the air right in front of the blade that is moved... but the general expression (showing the linear relationship with density and area, the quadratic relationship with velocity, and the dependence on the square of the attack angle) looks just like the one you quoted. Of course when $\theta$ (or $\phi$, in the expression you quoted) gets too large, the air flow will "stall" and the force will get smaller, not larger. That shows the limitation of this simplistic approach (which only works over a limited range of angles and velocities, and with all the other simplifications already mentioned).
A: Simplify, simplify, simplify.
If you throw something heavy, you have to push on it to do so.
If you throw some air (which is heavy) downward, you have to push on it, and that push pushes you up. That's lift.
OK, consider a simple wing.
Air comes at it, and it pushes that air downward (because the wing is slanted).
The speed of that downward velocity, times the mass of the air being thrown (per second), is the force it takes to do that pushing - the lift.
Here is my favorite explanation of all this.
After you fully grasp that (not before), then limber up your algebra and make it more complicated.
The wing is a certain size, and travels in a circle, etc.
