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4hv.org :: Forums :: General Science and Electronics
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Novel flying machines

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Dr. Slack
Mon Jun 23 2014, 01:16PM
Dr. Slack Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659
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

Try LibreOffice, OpenOffice didn't get onging development for some reason and forked into LO. That will open it.
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BigBad
Mon Jun 23 2014, 07:54PM
BigBad Registered Member #2529 Joined: Thu Dec 10 2009, 02:43AM
Location:
Posts: 600
I didn't have any problem with OpenOffice4.1.0
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Ash Small
Mon Jun 23 2014, 07:57PM
Ash Small Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245
Uspring wrote ...

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.




This it what one would presumably call a 'first order approximation'. I argue that all of the losses come from some form of drag, but it is just a question of semantics I think. You can probably identify several 'distinct' types of loss if you were so inclined, but it all comes down to the resistance of the air to being accelerated, or, in other words, it's viscosity.

I've drawn a quick graph. I assume the x axis (A) is m^2 and the y axis (v^3) is m/s.

I'm not mathematically minded, but I'll try and answer any further questions.


1403553463 3414 FT162858 Graph


EDIT: Actually I'm still trying to work out what the graph means wink

(if in doubt, draw a picture!!)

(x axis is 0-10, y axis is 0-100)

I'll sort out the PPT thing later, thanks for the tips.
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Patrick
Mon Jun 23 2014, 08:23PM
Patrick Registered Member #2431 Joined: Tue Oct 13 2009, 09:47PM
Location: Chico, CA. USA
Posts: 5639
im in the middle of a tri-blade experiment, results to follow shortly.
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Uspring
Tue Jun 24 2014, 09:19AM
Uspring Registered Member #3988 Joined: Thu Jul 07 2011, 03:25PM
Location:
Posts: 711
Ash Small wrote:
I argue that all of the losses come from some form of drag, but it is just a question of semantics I think. You can probably identify several 'distinct' types of loss if you were so inclined, but it all comes down to the resistance of the air to being accelerated, or, in other words, it's viscosity.
All of the props energy ends up as kinetic energy of air. Some part of it is useful, i.e. the downward air velocity, which results in an upward thrust. Other energies, e.g. the swirl around the blades and a helical component of the downward stream don't contribute to the upward thrust and are therefore regarded as losses.
Being picky: the resistance of air being accelerated is due to its mass, not its viscosity.
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Ash Small
Tue Jun 24 2014, 10:55AM
Ash Small Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245
Uspring wrote ...

Ash Small wrote:
I argue that all of the losses come from some form of drag, but it is just a question of semantics I think. You can probably identify several 'distinct' types of loss if you were so inclined, but it all comes down to the resistance of the air to being accelerated, or, in other words, it's viscosity.
All of the props energy ends up as kinetic energy of air. Some part of it is useful, i.e. the downward air velocity, which results in an upward thrust. Other energies, e.g. the swirl around the blades and a helical component of the downward stream don't contribute to the upward thrust and are therefore regarded as losses.
Being picky: the resistance of air being accelerated is due to its mass, not its viscosity.


All of these losses pretty much follow the same law, ie the 1/2A x v^3 thing, which seems to be common through all of the propeller efficiency/drag stuff.

I think it is something to do with viscosity, especially as the drag forces increase. I seem to remember getting a similar graph when dropping different sized ball bearings into a tube of oil as a student.

There is certainly more than one way to analyse any problem. wink
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Uspring
Tue Jun 24 2014, 11:34AM
Uspring Registered Member #3988 Joined: Thu Jul 07 2011, 03:25PM
Location:
Posts: 711
All of these losses pretty much follow the same law, ie the 1/2A x v^3 thing, which seems to be common through all of the propeller efficiency/drag stuff.
I think that we have a terminology problem. What in terms of the v^3 law the drag is for e.g. a car, is the thrust for a propeller. For the case of the car I would call it loss, but not for the prop.

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Ash Small
Tue Jun 24 2014, 11:39AM
Ash Small Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245
But does it not also equal the 'drag' losses from the periphery of the accelerated column of air?

These two 'forces' are equal and opposite, surely?
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Uspring
Tue Jun 24 2014, 01:19PM
Uspring Registered Member #3988 Joined: Thu Jul 07 2011, 03:25PM
Location:
Posts: 711
But does it not also equal the 'drag' losses from the periphery of the accelerated column of air?
The air column will eventually be slowed and stopped by the surrounding air. But that is not relevant to the thrust the prop generates.

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BigBad
Tue Jun 24 2014, 01:59PM
BigBad Registered Member #2529 Joined: Thu Dec 10 2009, 02:43AM
Location:
Posts: 600
It is somewhat relevant, depending on the Reynolds number.

The Reynolds number for a small drone is going to be a lot lower than a full size helicopter.

The lower Reynolds number means that viscosity is more significant than normal; model aircraft are designed slightly differently to full size aircraft for this reason.

Ash Small wrote ...

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:

Link2

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

Well, yeah, all aircraft are a bunch of compromises flying in close proximity, but I think you're completely overstressing the tip loss thing.

Ash Small wrote ...

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.
It gives the tip losses as about 3%.
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