# Integral? Average Velocity of a Spinning Propeller

1. Jul 21, 2012

### jaguar7

I'm trying to find the average velocity of a spinning propeller.

v(r) = RPM C(r) / t ; C(r) = 2πr; t=(1 min)
v(r) = RPM 2πr / (1 min)

I'm not quite sure what to do next. I need the answer in units of velocity, but if I integrate that last equation, I won't get units of velocity.

My rough guess at a next step would be:

lim(n→∞) (Ʃ(from r=0 to r_f) RPM 2∏r / (1 min)) / n
where "r_f" would be the "final radius," the "radius" being the "distance from the center"

===

For anyone who wants to know what I'm trying to do, exactly, I'm trying to plug a value for velocity (for a spinning propeller) into an equation for lift - a force.

If you want to see the math I did for that, it involves a bit of physics, but here;

We start with a "paddle in a river" analogy. A paddle of area, A, is held in a river, with water moving at velocity, v, the paddle held at an angle, a, to the direction of the motion of the water. The water has density, D.

What is the force exerted on the paddle by the water?

m=mass, p=momentum, t=time, F=force, D=density, V=volume, v=velocity, x=distance(in the direction of the water's velocity), A=Area, and a=acceleration in "F=ma" and a=angle in "sin(a)"

AvD = AxD/t = VD/t = m/t
∴ Av^2D = mv/t = Δp/Δt = F
Note momentum (p) = mv, so Δp/Δt = mΔv/Δt = ma = F

So (adding in sin(a), using the principle of vector components),
F_Lift = A v^2 D sin(a)
Note: this equation does not include a friction coefficient or any other coefficient to account for losses. You could add a multiplied by .75 or so or whatever to the end to account for losses.

Again, though, however, I'm on the step after that, trying to get the value for the velocity (of a spinning propeller) to plug into that equation.

Thanks,

Last edited: Jul 21, 2012
2. Jul 21, 2012

### Ibix

You don't want the average velocity, you want the total force. Say that the lift from the bit of blade between r and r+dr is f(r)dr=w(r)dr v2D sin a, where w(r) is the total width of blade at radius r (for a three bladed design with square blades of width w0, w(r)=3w0). Then sub in your expression for v in terms of r and integrate. If w(r) is non-trivial, I suggest a polynomial approximation, since your attitude to frictional loss suggests to me yot aren't looking for forty decimal places of precision.

Last edited: Jul 21, 2012
3. Jul 21, 2012

### HallsofIvy

Ibix has already told you that you don't need "average velocity".

But if you did, yes, just integrating, with respect to time, say, will give the wrong units- integrating with respect to time is equivalent to multiplying by as "time" and to get back to the correct units you have to divide by a "time".

Specifically, the average value of a function, from a to b, is given by the
$$\frac{\int_a^b f(x)dx}{b- a}$$
It is the division by b- a that corrects the units.

4. Jul 21, 2012

### jaguar7

Ibix: if wedo this, like you say:
F[v(r)] = Av^2(r)Dsin(a)
= wdr [RPM 2∏r / (1 min)]^2 D sin(a),
we would still have force on the left hand side. Integrating force would still give us incorrect units... we'd need to do something to fix or accommodate for that... I can't think of what to do though...

HallofIvy: I don't think we'd be integrating with respect to time though. We'd be integrating with respect to radius (or distance from the center), right?
For reference:
lim(n→∞) (Ʃ(from r=0 to r_f) RPM 2∏dr / (1 min)) / n (Note: this was an attempt at representing an "average," and I also exchanged dr for r, which I think would be correct)
So then you just divide it by r_f, then (r_f - 0)?
I verified from the wikipedia page for "average" and this looks like a method that should work.
If we use that method, we do this:
v_avg = 1/(r_f) (integral from r=0 to r_f): v(r)dr
= 1/r_f (integral from 0 to r_f): RPM 2 π r dr / (1 min)
= RPM 2 ∏ / [r_f (1 min)] (integral from 0 to r_f): r dr
= RPM π r / (1 min)
substituting "v = RPM π r / (1 min)" into the equation for force,
F = A [RPM ∏ r / (1 min)]^2 D sin(a)
I suppose we could "simplify" the RPM / (1 min) to generic angular velocity, as well, but RPMs are much more common (and useful) units to have there, I think.

I think Ibix's method has potential too, and it would be really great to have two methods, to verify the answer. But if we integrate that, I think we'd get units of work - not force. Maybe we could also divide by r_final... not really sure...

Last edited: Jul 22, 2012
5. Jul 22, 2012

### Ibix

Stick w in as a constant (units are m) and do the integral. See what the units are. You might be surprised.

The point is that it is not the integration that gives you the extra length dimension, it is multiplying by dr. I replaced A (an area) by wdr (an area) so the units are correct. Integration is just a sum. As HallsOfIvy tried to explain, you can divide your result by the integration range (the radius of the prop, in this case) to recover the original units. But, as we both said, you don't want to do this.

You do need to do something like what I proposed, which is to weight the velocity by the amount of propellor at that velocity. Otherwise, your calculation will tell you that taking thd blades off the propellor and fitting them outside-end-in would be just as effective as the right way round. Put another way, which is the more powerful propellor: three narrow radial spokes with vanes on the end, or three vanes at the boss with thin spokes on the end? Your method tells you there's no difference, since A is the same and $\int vdr/R$ is the same. My method gives the correct answer, since all the wide bits are where the velocity is high in one case, and where the velochty is low in the other.

Last edited: Jul 22, 2012
6. Jul 22, 2012

### jaguar7

Yeah, I need width as a function of r, since width isn't going to be constant. Which also means I can't use average velocity. You're right.

But I'm telling, you, dude - the integral of force with respect to a distance (r) is work. I probably shouldn't disagree or whatever because I don't want to start an argument. Everything you're saying is valuable, and I shouldn't argue, and honestly I am thankful for your response, overall, but that's the truth. It won't work. You need to divide by a distance before you integrate, or something.

HallsofIvy never said "you don't want to [divide by the integration range]."

If you replace A by w(r)dr, you still have an r where you sub v(r) for RPM 2πr / (1 min). When you integrate the whole equation, "w(r)dr [RPM 2∏r / (1 min)]^2 D sin(a)," you would get work - there is still an r variable where we replaced v(r) with "RPM 2∏r / (1 min)" so when you integrate you'll get an extra distance value - it will be r^2 - just like when you integrate velocity with respect to time you get distance, or you do the derivative of momentum with the respect to time, you get force. F*x = Work. v*t=x. It won't work. You cannot expect to integrate force with respect to radius and get an answer in units of force. An integral is not just a sum. It's the sum of an area of two variables. When you integrate force with respect to a distance, you think of the graph F compared to x, take the sum of the infinitesimals of F*x which gives units of work.

I think we need to divide the F(r) equation [with the w(r)] by an the range for r... that's just a guess though. But what you (seem to be) propose(ing) will not work, as I've explained.

Again, I probably shouldn't argue, and you have a great point with the w(r) being necessary, since the width won't be constant.

Thanks.

I guess the question at hand is if you integrate F/d (units of m/t^2), you'll get force, and will it be accurate in this case?
We have the following equation for force:
F[w(r), v(r)] = w(r) dr [v(r)]^2 D sin(a)
v(r) = [RPM 2∏r / (1 min)]

Can we just say w(r) = kr, for now? I don't know what kind of blades would be used, nor their function of width, yet.

Last edited: Jul 22, 2012
7. Jul 23, 2012

### Ibix

I made a mistake in my first post. I should have written that you should integrate:$$dF=w(r)dr v^2D\sin(a)$$and then you'd have been happy, I suspect. Essentially, what I called f(r) is not a force, it is a force per unit radius, and integrating it gets you back to a force. Do the integral and see what happens.

No - (s)he said
which is the point I was trying to make.

Units are:
w - m
dr - m
RPM/1min - s-1
r - m
D - kgm-3
That makes units of m*m*(m*s-1)2*kgm-3=kgms-2=N. You are correct that the function gains a factor of r when you integrate - but the dr disappears. So the units are the same.

Please do keep arguing. I think I'm right in this case, but I certainly am not always; neither of us will learn anything if we don't discuss our different understandings.

As long as k<2∏, what you are describing is pie-slice shaped blades. If k>2∏, the blades overlap. If that seems plausible to you, run with it.

8. Jul 24, 2012

### jaguar7

Okay, my bad. Thank you. ^_^

Well, moving forward; we might as well make "sin(a)" "sin[a(r)]" since propeller blades always have the element of "twist."

After integrating, and factoring out the constants, that leaves us with:
4∏^2(RPM)^2D/min^2 ∫(0 to r_f): r^2 w(r) sin[a(r)] dr

...which is a pretty complex equation... I guess I'll need to figure out the functions for twist and for width. I was hoping to just get an equation, and plug in a value for force, and then find out how much total area and RPMs I need. You're right, of course; propeller blades aren't constant width (or angle).

Maybe the previous equation (using average) would be useful when shopping or making a rough design for a propeller, and then we could use the "F = 4∏^2(RPM)^2D/min^2 ∫(0 to r): r^2 w(r) sin[a(r)] dr" later on, replacing the integral starting range accounting for the prop hub, and adding in a coefficient for losses.

Any thoughts, feedback, suggestions, corrections are more than welcome.