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Patrick

Tue Jun 17 2014, 10:05PM

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

Ash Small wrote ...

Patrick wrote ...

like the tupulev bear bomber, that would wipe the capitalist filth from the earth.

Sorry, I didn't make myself clear. I meant like these:

Patrick wrote ...

heres what im thinking about:

and

but with contra rotating props? and 'could you please provide some more detail?

Your post didn't make it clear that you were still considering contra-rotating props or not.

It does sound like a good idea to me.

(Sorry about the double post)

EDIT:

Yep, seems to give lowest loading in a surface area.

Ash Small

Wed Jun 18 2014, 12:52AM

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

Yep.

BigBad

Thu Jun 19 2014, 10:25PM

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

This guy used a 20Ah battery with 17 inch props on a quad and got over an hour flight time:

(figures are at the end of the video)

That's about twice the prop area than 12 inch props, and three times the battery, to get an hour flight; sounds about right.

Conundrum

Sat Jun 21 2014, 02:07PM

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

Hmm. I can't wait for "Mythbusters: Prison Breakout Special"... Escape via drone?! Seriously, this totally needs doing on a Bond film and soon.

AAaaaand in other news,

BigBad

Sat Jun 21 2014, 06:36PM

Registered Member #2529 Joined: Thu Dec 10 2009, 02:43AM
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I want to be picked up and carried around by a cloud of drones, even if I'm not escaping from a max security prison.

Patrick

Sun Jun 22 2014, 12:03AM

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

There is a new Ebola outbreak, monitoring the fruit bat population and their access to pig farms could be done with some more advanced instrumentation (years from now.) For example, 20,000 drones 10 cm in diameter, one time use, flown by 100 pilots for 25 medical and PhD experts could be seeded over the relevant areas. landing and taking off o a minute by minute basis.

The use of a single "bluefin 21" sub, to search for the maylasian airliner, seems like a child randomly scribbling with crayon, when Michelangelo would be using a brush. But maybe they only had one on hand.

In any case, fleets of drones are the future for practical application, not these single one off toys.

I do have concerns with landmine and AOL CD after use issues...

Ash Small

Sun Jun 22 2014, 08:25PM

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

OK, so maybe I will try to do some maths here after all. Power consumed is P = 1/2 * rho * A * vÂ³

which is half air density times area times velocity cubed, which shows that it's the velocity (presumably the air column velocity in this case) multiplied by half the area (disc area, presumably). I'll assume Rho to be a constant for now).

By far the most dominant factor here is the 'velocity cubed' term, which shows that any increase in air velocity is going to have a 'very negative' effect on efficiency, and a 'very large' effect on power consumed.

This is also very similar to the 'drag coefficient' equation, here: P = (1/2)(rho)(VÂ³)(A)(Cd) Where, Cd = Coefficient of Drag.

I assume in this case, because it is a column of air that is being acceleraled, that the drag coefficient of a column of air is equal to one.

These two equations then become exactly the same, which implies that the peripheral losses at the interface between the moving column of air and the surrounding, stationary, air completely dominate all of the losses from the system.

Now, decreasing the velocity of the column of air is exactly the same as reducing the disc loading. Both result in a larger, slower turning prop and an increase in efficiency due to a lower air velocity.

BigBad

Sun Jun 22 2014, 10:21PM

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

I'm not sure the disadvantage is quite as big as you think it is, otherwise twin rotor helicopters like the Chinook wouldn't exist.

Tip losses are treated fairly well here:

Ash Small

Mon Jun 23 2014, 09:22AM

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

BigBad wrote ...

I'm not sure the disadvantage is quite as big as you think it is, otherwise twin rotor helicopters like the Chinook wouldn't exist.

Tip losses are treated fairly well here:

Personally, I think the Chinook is a compromise for a number of reasons.

From what I remember, the blades on a Chinook overlap a bit (the rear prop is higher), so you end up with what approximates to one oval shaped column of air. Basically, if you reach the limits as far as one prop goes, you can halve the disc loading by adding another. There are other considerations. I do keep pointing out that it's a trade off between manouverability and efficiency. Counter-rotating props have other advantages too.

The maths does say that the 'propeller equation' and the 'drag equation' are identical, for all intents and purposes.

EDIT: Can't open PPT files, I boycott Microsoft software as much as I'm able to. Open Office doesn't seem to open them.

Uspring

Mon Jun 23 2014, 10:40AM

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

Ash Small wrote:

These two equations then become exactly the same, which implies that the peripheral losses at the interface between the moving column of air and the surrounding, stationary, air completely dominate all of the losses from the system.

These equations describe the kinetic energy that is contained in a moving cylinder of air. For a propeller, the air has to be accelerated in order to generate thrust, so it is not a "lossy" mechanism, but the intent of the prop. Not all of the power put into the prop is converted into air energy, so one usually defines a props efficiency by the ratio of air power to prop power.

In the case of air drag on a car, the idea behind the equation is, that a column of air in front of the car is accelerated to the speed of e.g. the wind shield. So the power lost by the car is the same as the power needed to give the air column its speed. From this a drag force can be calculated.

All of this has nothing to do with vortices around the prop tips. The equations quoted describe an ideal behaviour, where such effects are not taken into account. They can accounted for by adding some fudge factor to the equations.

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