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Patrick

Fri Jun 27 2014, 04:26PM

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

made another flight, it looks like a 8+ minute machine. thats with 10x4.5 multrirotor props. So, im thinking 12 or 14 inch props maybe the practical limit as far as diameter.

All-up-mass (AUW) = 1.61 kg
battery mass = 0.51 kg
unlaiden mass = 1.1 kg
lets say this equates to 8 min of flight.

Uspring wrote ...

this for 30 minutes or so of flight?

Flight time scales up linearly with battery size only if the battery mass is an insignificant part of total aircraft mass. Generally you'll have a time of flight dependency like this:

TOF ~ mb / (ma+mb)^3/2

ma being the aircraft bare mass and mb the mass of the battery. This will go through a max at mb = 2*ma. What's the mass of the copter w/o Lipos?

where did you get the 3/2 exponent? i only get 8 minutes if i use 3/1.6 exponent... (1.875) otherwise i think you have it close to my personal experience. so if i double the battery mass, it doesnt look good.that presumes i did the math right, im awfful tired...

Ash Small

Fri Jun 27 2014, 05:30PM

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

I need to draw another graph incorporating the thrust equation (F = rho * A * vÂ²) and the power equation (P = 1/2 * rho * A * vÂ³).

It seems obvious to me that increasing A and decreasing v leads to significant increases in efficiency, but I want to try to quantify this with a graph. (I tried plotting A against v^3 earlier, but I don't think I had the maths correct enough to tell me much, I need to incorporate the thrust equation, too.)

I'm currently thinking that ToF can be significantly increased by mounting the three props of an assymetric tri-copter on the 'legs' of a 'Y' frame, and mounting the motors, batteries, etc. at the intersection, and using reduction belt drives to turn much larger props than Patrick is currently considering.

It will probably take me all weekend to 'get my head round' the maths, though.

I'm convinced that the losses can be reduced by at least an order of magnitude without increasing rotor size to much more than ~1 metre diameter.

Patrick

Fri Jun 27 2014, 09:33PM

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

this is what i was planing...

torsional couplers, look at the melted one (Romanovs work) i dont know where to get these...
the ebay ones for 6.35mm to 5mm look like theyed melt too.

on this one we see the all metal type of coupler, not sure if these are meant for this purpose, or for stepper motors and such.

Uspring

Sat Jun 28 2014, 10:11AM

Registered Member #3988 Joined: Thu Jul 07 2011, 03:25PM
Location:
Posts: 711

All-up-mass (AUW) = 1.61 kg
battery mass = 0.51 kg
unlaiden mass = 1.1 kg
lets say this equates to 8 min of flight.

Here's a table of expected flight times for addittional battery packs:
The first comes from your measurement, the others are extrapolation.

1 Pack (0.51kg) 8min
2 Packs 10.6 min
3 Packs 11.5 min
4 Packs 11.7 min
5 Packs 11.7 min
6 Packs 11.5 min

where did you get the 3/2 exponent?

This is from the equation relating thrust to power consumption:

F = eff * (2 * rho * A * P^2)^(1/3)
so the proportionalities
F ~ P^(2/3) or
P ~ F^(3/2)
hold. Since we hover, we have
F ~ ma+mb
for the flight duration we have
TOF ~ 1/P

Put this all together and there is the to the power of 3/2 dependence.

Ash Small

Sat Jun 28 2014, 12:10PM

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

If I ignore Rho for the time being, F=Av^2 and P=1/2Av^3.

There's something I'm missing, I think.

Udo wrote:"F = eff * (2 * rho * A * P^2)^(1/3)
so the proportionalities
F ~ P^(2/3) or
P ~ F^(3/2)
hold."

I don't see how you can relate F and P in this way. I'm probably missing something.

Surely power is dependant on A? The smaller A is, the more power is needed to overcome additional losses in order to achieve the same force?

I'm trying to reach the point where I can plot efficiency against A, or power required to achieve required thrust for varying areas.

I don't see how the relationship between P and F can be as simple as suggested. (unless A is constant)

EDIT: I found some more equations,

Power (WATTS)=P(in.) X D(in.)^4 X RPM^3 X 5.33 X 10^-15

Thrust (oz.)=P(in.) X D(in.)^3 X RPM^2 X 10^-10

Apparently they are called the 'Abbot equations'

EDIT: Also managed to get this spreadsheet based on the Abbot equations. I don't know how accurate or reliable it is yet. I had to rename it as a 'text file' to upload it here, you need to change the file extension to .xls before you can open it. I did this using 'cmd' and copying it .

]propeller.txt[/file]

EDIT: Also managed to download another spreadsheet from the same page as above, this one was posted by Bruce Abbot a bit further down the page. I'm not sure if it's much use for what I'm trying to do, though.

EDIT: I think I might be getting somewhere with transposition,

F/A=v^2

P/A=(v^3)/2

still thinking about it.

Conundrum

Sat Jun 28 2014, 12:31PM

Registered Member #96 Joined: Thu Feb 09 2006, 05:37PM
Location: CI, Earth
Posts: 4062

Is there a way to use a lighter pack but with supercapacitors, ie lift-slow descent alternately?

BigBad

Sat Jun 28 2014, 01:23PM

Registered Member #2529 Joined: Thu Dec 10 2009, 02:43AM
Location:
Posts: 600

no, not really. This is strictly an energy/energy efficiency/total weight thing, not a power thing. supercapacitors have rotten energy density by weight; but are fairly high power.

Uspring

Sat Jun 28 2014, 03:37PM

Registered Member #3988 Joined: Thu Jul 07 2011, 03:25PM
Location:
Posts: 711

I don't see how you can relate F and P in this way. I'm probably missing something.

The relation holds if you keep the prop area A constant. The ~ symbol is meant as "proportional to" not as "about the same". As a physicist I use the equal sign for "about the same"

Ash Small

Sat Jun 28 2014, 04:39PM

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

Uspring wrote ...

I don't see how you can relate F and P in this way. I'm probably missing something.

The relation holds if you keep the prop area A constant. The ~ symbol is meant as "proportional to" not as "about the same". As a physicist I use the equal sign for "about the same"

Thanks

Patrick

Sat Jun 28 2014, 05:51PM

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

Uspring wrote ...

All-up-mass (AUW) = 1.61 kg
battery mass = 0.51 kg
unlaiden mass = 1.1 kg
lets say this equates to 8 min of flight.

As expected, im adding more and more mass for less and less flight time.

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