#### GregP

##### Captain

Tip speed is not D * n. Tangential tip speed is r * omega (radians/ second). If r is in feet, tangential velocity is feet per second. If r is meters, then it is in meters per second. But, you have to factor in forward velocity in the same units, too. Prop tip speed = square root [(forward velocity)^2 + (tangential velocity)^2].Allow me to throw in a few considerations. Assume we're designing a high-speed propeller aircraft from scratch. A single constant-speed propeller is a given. Our variables are propeller diameter, rpm (we have a gearbox that we can design to optimize propeller performance for our engine), number of blades, blade chord, blade planform, and angle of attack. We want to go as fast as possible, because...zoom and boom.

A designer would typically start with diameter, since whatever is choses has to match with the rest of the design concept. A good rule of thumb is:

D = 15.24 * Power^(0.2)/(n^(0.6))

D is in meters, power in horsepower, n is rotations per second. I usually don't mix units like that, but it is what it is.

Going fast requires MOAR POWER!, so that will probably be constrained by whatever the guys in the engine department come up with. Let's just say they have something of P horsepower, and we'll use that. Plug whatever number P is into that equation, and you can develop a curve of D vs. n.

Ah, but we all remember tip speed has to be kept below Mach 1 (in real world terms, maybe 950 ft/sec). Tip speed at any given velocity is determined by D * n...we can't do anything about the high speed we're flying at, but we want to keep D * n as low as possible to keep the tip speed down.

Let's throw some numbers in. D * n^(0.6) = 15,240

D = 1, then n = 9,368,000

D*n = 9,368,000. A little high. (I suspect the units are wrong, maybe it should be rpm instead of rps?)

D = 2, then n = 2,951,000

D*n = 5,901,000. Better.

D = 3, then n = 1,501,000

D*n = 4,504,000

I think you can see where this is going. We want D to be really big and n to be really slow. We'll probably hit some practical limit before we get to a 50-foot propeller, but at first glance, it seems to make sense to design the thing around having as big a prop as possible. It is here that we must introduce "Actuator Disk Theory" or as it is sometimes known, "Prop Momentum Theory".

Imagine that our aircraft has a magic disk, a hoop that accelerates air backward like a jet engine. We can make it as big or small as we like, and pump P amount of power to it, and it will take that energy and give 100% of that energy to all the air that passes through it. What you'll find is that if you make the magic disk really small, it really shoots the air back hard in a super powerful jet, and if you make it huge, it barely puffs an enormous amount of air backward at the speed of a gentle breeze.

And here's the big catch. The energy you give to the air is determined by the kinetic energy formula, KE = 1/2 * M * V^2, but the useful thrust that pushes the airplane forward is given by the momentum formula, P = MV.

So, should you make the hoop small, and shoot the air out fast (really big V) or make it big and shoot the air out slow (very small V)?

If you have INFINITE power, it doesn't matter. But if you are LIMITED on power, you can get more momentum (MV) with less kinetic energy (1/2MV^2) by pushing it out slow. Big disk, here we come!

You've probably figured out that our magic disk is an ideal propeller, one that has no real-world losses from friction or turbulence or anything like that. Now, here's the killer, a graph of the efficiency of ideal propellers as a function of diameter:

View attachment 702169

Ah. So that looks like we want to make the prop big, but beyond a certain point, it doesn't do any good to get much bigger. And remember, these are imaginary, perfect propeller losses, not real-world losses, which are worse. I guess we might not be stealing that big prop from a Tupolev Tu-95 after all.

So, with a typical WWII-era aircraft engine, we do some design studies, and we get a diameter about equal to what flew on typical WWII-era aircraft. Engine location, fuselage length, landing gear are all designed so that this diameter fits with whatever clearance factor whoever makes that decision likes. We then design a gearbox that keeps the tip speed somewhere between, oh, let's say 750 fps and 950 fps.

Our next problem is ensuring that whatever prop we design can soak up that much horsepower and convert it to useful thrust. Blade angle is taken care of for us, it will be a function of the NACA section chosen for the prop, and they all top out somewhere in the neighborhood of 45 degrees. Go past that, and you'll still make thrust, but not as efficiently, and remember, we're limited to the horsepower the engine guys give us. So, we figure out what angle is the top of the thrust curve for our prop section, and base our decision around that angle at our highest operational speed.

That leaves chord length and number of blades. A long, skinny blade is more efficient than a short, fat one, and fewer blades is less efficient than more blades. So, we start with two absurdly fat blades (big chord, although for the same section the prop will get thicker as well). We then compare with three thinner ones, then four even thinner ones, each time picking the smallest chord that just gets that much power to the airstream at the chosen RPM. We run the design studies, and whichever one wins is our design. I think you can see it's a function of power and diameter. The smaller a diameter for a given power, the more blades we'll need to be supremely efficient.

Why is smaller diameter faster (more efficient?) Because an existing plane (especially an air racer) can't just gear to whatever speed they'd like. The engine has a power curve, and it makes more power at a certain RPM, and that RPM may not be the same RPM that the aircraft was designed for. Sometimes, smaller is faster, sometimes, bigger is faster, rarely is stock faster after you make a lot of engine mods. If you are designing a turboprop and have budget to re-gear, then smaller will probably not be better...unless you add so much power that you need to add a blade, anyway.

Your propeller tips will be rounded for the same reason Spitfire wings were elliptical...it is even more important that you translate every iota of engine power to useful thrust. Why do you see aircraft with square tips? Because they've been re-engined, and it is either more efficient, or nearly as efficient and much cheaper to refit with squared-off blades than it would be to add blades for the higher power engine. Or maybe it was a new design, and it was cheaper to borrow a prop already in production for another aircraft. But you would never design one from scratch that way, it would be a compromise of some sort, probably to maintain ground clearance in a refit.

Just a few thoughts on how all this comes together.

Lets say you are at 25,000 feet going 450 mph in some wonder chariot. You have an 11 foot diameter propeller and your engine is turning 3,000 rpm.

The prop reduction gear is 0.42 : 1.

1) 450 mph = 660 feet/ second.

2) Propeller rpm = 3,000* 0.42 = 1,260 rpm, so Omega = 131.947 radians/ second.

3) Tangential speed = (11 / 2) * 131.947 = 725.708 feet/ second.

4) Prop tip speed = square root (660)^2 + (725.708)^2) = 980.926 feet/ second.

5) At 25,000 feet on a standard day, the speed of sound is 1,014.3 feet/ second.

6) So, the tip speed is (980.926 / 1,014.3)= Mach 0.97. Not very likely since the tip speed is so close to M1.

I'd feel a LOT better about the hypothetical situation if the tip speed were around Mach 0.87 or so. For our aircraft /prop / engine combination, our forward speed would have to drop to 342.5 mph to get the tip speed to M 0.87.

Alternately, we could still make Mach 0.92 if everything was the same except our prop was 10 feet in diameter and we could put up with a tip speed of M0.92.

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