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Patrick

Thu May 29 2014, 07:46AM

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

Dr. Slack wrote ...

Now for Mk2, if you mounted the motor on a bearing, and resisted the rotation with some means of measuring torque... ....you could measure true torque/power at the prop, and you'd have a motor dynamometer as well.

hell. i should have thought of that.

there is a super dooper important PDF, but i cant get it to save in the attachments section.

hers the link:

the velocity and pressure diamgrams are important. i may throw flour into the propeller to visualize the exhaust stream.

Patrick

Sun Jun 01 2014, 05:46AM

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

If 'm dot' is a derivative, could it also be written as dm/dt ?

Uspring

Sun Jun 01 2014, 08:58AM

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

Yes.

BigBad

Sun Jun 01 2014, 02:06PM

Registered Member #2529 Joined: Thu Dec 10 2009, 02:43AM
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m dot is Newton's notation; he only knew how to differentiate wrt time

dm/dt is Leibnitz's notation, it's more general.

Ash Small

Mon Jun 02 2014, 09:45PM

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

Not only do I have helicopters flying past my windows most days (I live on a hill in a military training area), but recemtly they've been flying around my head. I think if you have a 'Chinook' configuration, but with two pairs of concentric props (four props altogether), you can do it just with VPP, without swashplates 'IF' you drive each prop from a separate motor.....still thinking....and drinking....

EDIT: It's possible to do without VPP as well, if you use electric motors.....four motors, four props, but two concentric pairs, in Chinook configuration.

Patrick

Sun Jun 08 2014, 07:19PM

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

im thinking about all these derivations, i dont think i need the square on the (Ve - Vh) part since im using the derivative of m. is this right?

Uspring

Tue Jun 10 2014, 10:17AM

Registered Member #3988 Joined: Thu Jul 07 2011, 03:25PM
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The first equation is Newtons law of motion.
The second one is wrong.
The third describes the thrust of e.g. propeller, which accelerates air at the rate dm/dt from velocity vh to ve. Think of the situation of pushing a mass m with the force F for a time t. Then the mass will have accelerated to the velocity v. The involved quantities are related by the equation

F*t = m*v

This equation describes the conservation of momentum. Divide both sides by t and you get:

F = (m/t) * v

This is basically your third equation for the special case, that the entrant air velocity vh is zero.

Ash Small

Tue Jun 10 2014, 02:10PM

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

I think the second one should be something like F=m(v2-v1)/t (it's a long time since I was a student).

BigBad

Tue Jun 10 2014, 02:10PM

Registered Member #2529 Joined: Thu Dec 10 2009, 02:43AM
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The m and the v are related, at high v_e, you are drawing more air through your rotor per second.

i.e.

m ~= rho v_e A

where rho is the air density and A is the disk area

but the reality is more complicated; some of the air is vortexing around in a circuit and not being accelerated as much, so giving less force; you could probably have some sort of fudge factor for that. However, due to viscosity you'll still get advantage from even that.

But the rho v_e A is probably a good start.

Basically:

F = m dv/dt

let's say that every second you're accelerating a mass m of air, from rest above the vehicle to V_e, as a jet of air below the vehicle:

F = m V_e

substitute for 'm'

F = (rho V_e A) V_e

rearrange:

F = rho A V_e^2

(n.b. you have to be a bit careful with those dm/dt terms, mass is conserved, so dm/dt is always technically 0!)

Ash Small

Tue Jun 10 2014, 04:57PM

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

BigBad wrote ...

but the reality is more complicated; some of the air is vortexing around in a circuit and not being accelerated as much, so giving less force; you could probably have some sort of fudge factor for that. However, due to viscosity you'll still get advantage from even that.

If BigBad is reffering to the peripheral vortex losses, these are much greater for a smaller, faster turning prop than they are for a larger, slower turning prop, assuming both are consuming the same power, for two reasons, firstly, the losses will be greater from the faster moving air, and secondly, the cross sectional area/circumferance ratio means that, for a larger, slower turning prop, a smaller percentage of the moving column is at the periphery, and experiencing losses.

You want to aim to accelerate the largest volume of air possible to a low velocity, as this is far more efficient than accelerating a small volume of air to a high velocity, due to these peripheral losses.

EDIT:The 'trade-off' is manouverability. Dr Spark's drone with 5" blades spinning at 25,000RPM demonstrates this exactly, very manouverable, but short flight times.

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