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Old Dec 09, 2007, 10:52 AM
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Question

Conversion factor2 to 3 blade prop


Hi all, need to know how to convert 2 bladed prop to 3 or 4 blades. Landing gear to short for bigger prop. Need to use a 3 or 4 blade prop. Need to know what the conversion factor is. Thanks In Advance, Rich
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Old Dec 09, 2007, 11:06 AM
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Here on this forum the Good Dr. Kiwi has tested some of the new GWS HD three-bladers. All the info you would ever need is most likely there. or here...
https://www.rcgroups.com/forums/showthread.php?t=781035
Last edited by 56S; Dec 09, 2007 at 12:28 PM. Reason: can't spell!
Old Dec 09, 2007, 11:25 AM
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I haven't yet accumulated enough data to determine what 2-blade prop is the exact equivalent (thrust/pitch speed) to a 3-blade prop. .... but I'm working on it!
Old Dec 09, 2007, 07:29 PM
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Propeller design theory says:
2 to 3 blades, reduce diameter by 10%,
2 to 4 blades reduce diameter by 15%,
pitch stays the same in both cases.

Steve
Old Dec 11, 2007, 01:15 AM
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aireze,

There is a very complicated formula for determining the load factor of a propeller, but in it's most simplistic form, for a 2-bladed prop, the load that a prop places on a motor is equal to the diameter cubed times the pitch or D x D x D x P. For a 2 bladed 10x6 prop, the load factor would be 10 x 10 x 10 x 6 or 6,000. For a 12x8 prop it would be 12 x 12 x 12 x 8 or 13,824.

The more complete formula, which takes the number of prop blades into account is D x D x D x P x Square root (N-1), where N = the number of prop blades. For a 2 bladed prop, the square root of (2-1) is the square root of 1 which is 1, so the term just drops out of the equation.

For a 3-bladed prop, the correction factor is the square root of (3-1) or the square root of 2, which is 1.414.

For a 4 bladed prop, the correction factor is the square root of 3, which is equal to 1.732

So if you have a 3-bladed 9x7 prop, then the load factor is 9 x 9 x 9 x 7 x 1.414, which is 7,216, and this would be roughly wquivilent to a 2-bladed 10x7 prop, which has a load factor of 7,000.

If you had a 4-bladed 12x7 prop, then the load factor would be 12 x 12 x 12 x 7 x 1.732 or 20,950 This would be roughly equivelent to a 2-bladed 14x8 prop, which has a load factor of 21,952.

In the end, if the load factor of 2 props is the same, you will get similar RPMs from the two props, and similar performance.

So there you go, hope that helps!

Lucien
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Old Dec 11, 2007, 09:05 AM
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Go down 1 in. diameter with the same pitch will put you in the ball park.

Gord.
Old Dec 11, 2007, 09:55 AM
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I've always used this formula: 2/x to the .25 power. 2 = number of blades of two bladed prop. This is a constant. X = number of blades you want (usually 3 or 4) The answer you get is a decimal. Multiply it by the diameter of the two bladed prop and your answer is the new prop. Use the same pitch.

To simplify; for a three blade, 2/3 raised to the .25 power = .903602 This is the two to three blade factor.

To go from a two to four blade, 2/4 raised to the .25 power = .8408964 This is the four blade factor.

2/5 to the .25 power = .7952707 5 blade

2/6 to the .25 power = .7598357 6 blade

So, if you have a 10-7 two blade and want a three blade: 10 x .903602 = 9.036 diameter. Round it off to 9, use same pitch, 9-7 three blade.

If you start with three blades and want to go to two, divide the three blade diameter by the same .903 factor to get the two blade diameter.

These factors are virtually identical to what Slipstick said.
Old Mar 31, 2008, 04:57 AM
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Thanks Lucien

Peter
Old Mar 31, 2008, 11:14 AM
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There cheap enough at www.gwsprops.com to try them out in all the sizes.

The Direct drive GWS shaped 3 blades are quite a bit differant in comparison amp draws than the SLO fly series 3 blades. Something you must consider.

I generally just drop down by 1" in the dia of the 2 blade as a starting point. I am running more 3 blades than 2 now days as the torq effects for same thrust ratings are reduced. I preffer only the Direct drive profile however.
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Old May 22, 2016, 03:57 PM
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Really appreciate the posts here guys learned alot!
Old May 31, 2016, 08:09 PM
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Quote:
The more complete formula, which takes the number of prop blades into account is D x D x D x P x Square root (N-1), where N = the number of prop blades. For a 2 bladed prop, the square root of (2-1) is the square root of 1 which is 1, so the term just drops out of the equation.
Interesting equation in it being simple, but something is not quite correct, as it implies a single blade has zero load factor..?
Old Sep 01, 2016, 04:58 PM
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sorry wrong thread


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