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Propeller Physics and VABs Program.

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Ash Small

Mon Dec 08 2014, 12:36PM

Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245

Patrick wrote ...

one guy is using pipe, then cutting it off to make his own props.

composite parts.

pipe cut props.

odd one.

Someone says just about any prop can be 70% or so efficient. Since were approaching 100% as an asymptote, eventually theres no advantage in going all out for that last 1%. (im tipsy so this was probly said already) So lets find out what the efficiency of the APC 10, 11, 12 and 13 MR's are, then we'll see how much margin is at stake.

APC has all these numbers, just waiting for re-posting them.

Well, I don't know how the bloke in the first link is getting 60PSI of vacuum. 15PSI is the limit of all conventional vacuum systems

I think we discussed laying up on a cylindrical form before. You still need to consider leading and trailing edges, etc.

I'm still of the opinion that all of those commercially available props are designed for much higher disc loadings than we want, so will be inefficient at low disc loadings.

I think we need to consider RPM next, once we know that we should be able to do some initial calculations on blade profile, angle of attack, etc.

Patrick

Mon Dec 08 2014, 06:19PM

Registered Member #2431 Joined: Tue Oct 13 2009, 09:47PM
Location: Chico, CA. USA
Posts: 5639

That "Test Pilot Mafia" guy... thats me

Various configurations. It seems they dont want CF props to see more than 22.2 V, the wood props are rated to 44.4 V at near 4,000RPM.

the best effciency from 19.4 g/W to 15.6 g/W.
this varies across 1,100 to 1,550 RPM.
From 50 to 75 % throttle, (this is where hover and simple maneuvering is done.)

The best efficiency i measured on my thrust stand was 8.7 grams. 19.4 is 2.2 times better per unit of power.

Pics of the U8. its a pancake motor, just not sure why these physical shapes are chosen, still studying this too.

in any case, as youve said low RPM at large diameter yields high effciency. the above props, ive been told "explosively liberate" themselves at 6,000RPM or so. (i dont know if they're thin at the hub or what)

Shrad

Mon Dec 08 2014, 06:55PM

Registered Member #3215 Joined: Sun Sept 19 2010, 08:42PM
Location:
Posts: 780

I just saw on Discovery that porsche 318 motors are pancake motors of that kind, so there should clearly be other advantages than form factor...

Ash Small

Mon Dec 08 2014, 10:31PM

Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245

Patrick, just got back after a long day, but on the subject of props disintegrating at high RPM, there are basically two factors. One is if they are designed for high disc loading they will be producing loads of thrust at high RPM. The other is that if designed for high disc loading they will be much heavier, and therefore the centripetal/centrifugal forces are also high.

I think a low disc loading prop will spin much faster, due to it's low mass and due to it not generating a lot of thrust.

I know I previously said 'a big prop at low RPM, but I should have said 'or a big prop at the same RPM with less cross section'. I think we want fairly high prop RPM for increased manouverability.

If we increase RPM, we can reduce mass and cross section, have less angle of attack, etc.

The only limit seems to be, apart from centripetal/centrifugal stuff, drag on the blades, but we can reduce turbulence behind the blade, and improve laminar flow.

Does this make any sense?

I'll have a look at the charts you posted tomorrow.

Andy

Tue Dec 09 2014, 02:29AM

Registered Member #4266 Joined: Fri Dec 16 2011, 03:15AM
Location:
Posts: 874

Apologies Ash,

You've not taken losses into account here, Andy, which increase as prop size decreases and as air speed increases.

The most efficient is always large diameter and low RPM, and it's pretty much exponential, very similar to the 'drag equation'.

Just doing some thinking and F = 1/2mv2 using double simposim position becomes

2m2v2, dropping back to or base shifting equals v3(velocity cubed) 6 = v*1/2
If we take power in watts used P = 1/2mv2, releatising that a system is equal and opposite and that opposite energy gets removed out of the system to a larger system, we have 2mv2*2, the opposite value(1/2) 4 = v

v3/v4 power is 75% effeiect ln(75%) = 25% loss with a increase in speed per step, 50% of 75% is 42.5% which is the effeinect of blades and props to convert and use energy in wind flow.

Patrick

Tue Dec 09 2014, 05:20AM

Registered Member #2431 Joined: Tue Oct 13 2009, 09:47PM
Location: Chico, CA. USA
Posts: 5639

Text files !

i found this math interesting :

]per3_10x45mr.txt[/file] multi-rotor

]per3_10x55mr.txt[/file] multi-rotor

]per3_11x38sf.txt[/file] slo-fly

]per3_12x38sf.txt[/file] slo-fly

]per3_12x45mr.txt[/file] multi-rotor

]per3_14x55mr.txt[/file] multi-rotor

if we look at 0 to 3 mph, we see most of the power being used as static force. Right?

Ash Small

Tue Dec 09 2014, 12:34PM

Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245

Patrick wrote ...

Text files !

i found this math interesting :

multi-rotor

multi-rotor

slo-fly

slo-fly

multi-rotor

multi-rotor

if we look at 0 to 3 mph, we see most of the power being used as static force. Right?

At first glance it looks like those figures are obtained in a wind tunnel, and they've measured power required for 2000RPM at various wind speeds. Is this how you interpreted it, or am I missing something?

Patrick

Tue Dec 09 2014, 04:29PM

Registered Member #2431 Joined: Tue Oct 13 2009, 09:47PM
Location: Chico, CA. USA
Posts: 5639

Ash Small wrote ...

At first glance it looks like those figures are obtained in a wind tunnel, and they've measured power required for 2000RPM at various wind speeds. Is this how you interpreted it, or am I missing something?

\i geuss i should have asked. dont know, duh.

Ash Small

Tue Dec 09 2014, 06:29PM

Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245

Patrick wrote ...

\i geuss i should have asked. dont know, duh.

Well, I think those datasheets are telling us thrust at various speeds at 2000RPM, but they do also tell us thrust and power at hover, so I suppose we can work out the efficiencies.

I've been reading up a bit on full size helicopters and found out that most have a tip speed of 700 feet per second, which equates to ~6000RPM for a 26 inch prop.

Maybe we should try some maths for a prop like this at,say, 5000RPM to make the maths simpler?

This would be maximum RPM, so RPM for hover should be ~2500RPM.

We know the thrust required for hover, so we should be able to get to a theoretical pitch.

Patrick

Tue Dec 09 2014, 09:56PM

Registered Member #2431 Joined: Tue Oct 13 2009, 09:47PM
Location: Chico, CA. USA
Posts: 5639

in a hover, we have 100% slip, right?

So how are we to figure on pitch? i keep thinking about static thrust and screw threads in wood. but my brain breaks...

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