Joined Jun 2012
Why does prop thrust increase with fourth (not third) power of diameter?
Quick theory question,
I'm pretty sure I understand how all different effects increase the size of the indices in the equation, but I can't think of what the last effect is.
As diameter increases,
To start off, prop size increases linearly with D, which linearly more prop doing the same thing, i.e. linearly more thrust therefore T∝ D
Because the blade isn't moving straight through the air, but rather is on a pivot, that added prop is going linearly faster - Aerodynamic forces scale with the square of speed (e.g. drag equation, lift equation), so that gives us two more powers: T∝ D*(D*D) = D^3
But all the equations say that thrust increases by the fourth power of diameter, which means either I am missing one modifier, or I am misunderstanding part of the last two. I can't think of what it is? Can anyone let me know? The only thing I can think of right now is that the equation is assuming the chord of the prop is increasing with its diameter, and that linearly increases thrust and gives us D^4.
Someone halp me learn D:
Joined Dec 2002
Pitch may be considered a function of Diameter.
If you take the standard lift equation
Lift (Thrust) = 1/2 X rho X Cl X V^2 X A
put in D where it belongs in the terms V, A and Cl, then integrate from the center to the blade tip it results in a D^4 term.
The extra term comes from the fact that the equation is assuming a constant pitch/diameter ratio.
If the pitch term is also considered, Diameter^3 is the correct term. Pitch may then be considered as a percentage of Diameter and the net result is a D^4 term.
For your interest, the equation below is for the thrust of a helicopter rotor.
Diameter and Chord are in inches. Lift (Thrust) is in ounces.
Lift (Thrust) = 0.0024(Rho) X Cl X (2 X Pi X RPM/60)^2 X (Diameter/24)^3 X (Chord/12) /3 X16
In this case the Diameter term is cubed because the Pitch term which may be expressed as a function of the Diameter is contained in the Lift Coefficient term Cl.
Thrust Power Derivation
Fundamentally, thrust is proportional to rpm squared times diameter squared >> T = rpm^2 * D^2. There’s a few constants that need to come along though: rho ()/density (temp & pressure effects), pi (), a prop constant, etc.
Most often the prop constant gets determined empirically, and is usually expressed in terms of Pitch and Diameter. The most common form for thrust though doesn’t use pitch, but rather diameter squared.
Using R as rpm from hence forth.
And then, most folks don’t talk about the fundamentals so what you’ll find written is >> T = k*D^4*R^2. A few folks write as suggested above T = k*P*D^3*R^2; not much difference when P and D are close in value, like a “square” prop 5x5.
Power is actually more important in understanding a prop’s performance in E-Flight, and is the parameter that the various Calc programs use to predict a configuration’s performance. Again, it is rarely shown in its fundamental form, being written typically as Watts = k*P*D^4*R^3.
By 1st principles power is velocity times thrust. Maybe you’ve noticed the real fundamental parameter by now – Velocity!
Thrust for any propulsion system (i.e. EDF, or rocket) written from 1st principles is k*rho*V^2*A^2 – velocity (e-flux) squared applied over an area squared.
A prop turns R into V >> V = k*P*R which leads to all of the above.
Pitch speed (velocity) is commonly calculated as P*R/1000 using pitch in inches; this is an approximation that neither applies a “k”, nor use the proper constants for unit conversion.
Multi bladed props are more efficient at turning R into V!
Multi bladed props are more efficient -- at turning R into V!
That’s probably what we all remember.
But, Newton actually stated that a force is equal to the time rate of change of momentum (mass times velocity) = d/dt mass * velocity – this difference is very important with respect to propulsion.
In Kinematics we think of F = m * a, where acceleration is d/dt velocity.
In fluid dynamics its d/dt (m), mass flow; so its mass flow * velocity – this is known as impulse.
The momentum theory shows that thrust is a function of area and velocity squared.
Thrust for any propulsion system (propeller, EDF, rocket, or Ion drive) written from 1st principles is k*rho*V^2*A^2 : velocity (e-flux) squared applied over an area squared.
Since most of those have round exit planes, thrust is a function of:
T = f(diameter squared & rpm squared) >> T = rpm^2 * D^2.
(Using R as rpm hence forth)
A prop turns R into V >> V = k*P*R .
Prop constants are most often determined empirically, and are usually expressed in terms of Pitch and Diameter. The most common form of the thrust equation seen around here doesn’t use pitch, but rather diameter squared.
What you’ll often find is >> T = k*D^4*R^2. A few folks write as suggested above T = k*P*D^3*R^2; not much difference when P and D are close in value, like a “square” prop 5x5.
By 1st principles power is velocity times thrust. Maybe you’ve noticed the real fundamental thrust parameter by now – Velocity!
Power is actually more important in understanding a prop’s performance in E-Flight, and is the parameter that the various Calc programs use to predict a configuration’s performance -- Watts = k*P*D^4*R^3.
Another look via fundamental units: Length, Mass & Time:
Velocity = L/T
Force = M*L/T/T
Power is M*L*L/T/T/T
Velocity times Thrust = L/T * M*L/T/T;
Which = M*L*L/T/T/T -- Power!
As a check, force divided by power leaves you with a parameter that has the fundamental units of T/L. = Velocity!
Note: efficiency, like Reynolds Number is a parameter w/o units -- it makes no sense if it has units; as is the case when using thrust/watts: F/P = (M*L/T/T) / (M*L*L/T/T/T) = 1/ (L/T); not w/o unit -- units of Velocity remain.
So we see velocity is key – what’s velocity of a prop – prop wash?
Pitch speed (velocity?) is commonly calculated as P*R/1056 using pitch in inches; this is an approximation as there’s no “k” applied to account for losses.
Is pitch speed anywhere close to prop wash velocity? – NO; not even close. PS – not related to or puts a limit on flight speed.
Though adding blades gets you closer -- multi bladed props are more efficient -- at turning R into V!
I measure velocity using a Kestrel 1000; it’s actually one of the easiest parameters to measure.
And, it is very useful.
Square it and you don’t need to measure thrust.
Cube it and you have Power Out.
Hmm…. cubed divided by Whattmeter Watts – Efficiency!
Joined Nov 2005
Joined Jun 2012
So in conclusion, some of you have mentioned it is due to pitch increasing for P/D to stay the same, and some have mentioned it is due to chord increasing as the whole prop scales up. Either way it is because the formula assumes more than just D changes. Unfortunately, both of the points are valid but there is only room for one. So I did some thinking and I figured out - actually there is room for both due to another effect:
Just as the average speed of the prop decreases as D increases because the new section of prop moves at linearly higher speed - the average AOA of the prop is actually decreasing as the new section of prop has linearly decreasing AOA (due to twist). This is important - another application being that the whole reason aero forces increase by speed squared is that objects moving through air at higher speed move both more air, and move it faster. We are no longer moving it faster if we are reducing the AOA as we increase the speed.
So, the effects of changing prop geometry and speed are:
1) Thrust from increased RPM, changes ∝ n^2
1) Thrust from Increased average speed due to diameter, changes ∝ D^2
2) Thrust from Increased length due to diameter, changes ∝ D
3) Thrust from Increased Chord, (not sure the symbol but calling it C) changes ∝ C. We assume this is proportional to diameter, therefore changes ∝ D.
4) Thrust from decreased average AOA, changes ∝ D^-1
5) Thrust from increased pitch, changes ∝ P.
So in total we get this:
Thrust ∝ N^2*D^2*D*D*D^-1*P = N^2*D^3*P
The only way we get the conventionally used N^2*D^4 relationship is if we, as Martin Mckinney pointed out, assume that P/D ratio is constant and thus P∝D, so we get:
Thrust ∝ N^2*D^3*D = N^2*D^4
Does this sound reasonable, everyone? I hope so because it makes me feel much better as this has been nagging at me for weeks if not months.
@Jrb, I read everyone's posts in bed this morning on my mobile because I couldn't wait to read them, and thought about this response in my head at the time. At the time, with my still-asleep brain, I couldn't really comprehend your post, so I haven't had a chance to read, digest, and respond to you yet. It looks interesting though. Just give me some time :P
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