• Announcements

    • Cap'n Refsmmat

      SFN Upgraded   07/22/17

      SFN has been upgraded to IPB version 4. View the announcement for more details, or to report any problems you're experiencing.
Sign in to follow this  
Followers 0
Capiert

How much Power do I need to levitate 1 kg?

41 posts in this topic

The reason I "had" thought the answer was 48 W/kg

was because

the power

P=F*va

is the force F (as weight Wt=m*g=1 kg*9.8 m/(s^2)=9.8 N)

multiplied by

the average speed

va=(vi+vf)/2=(0 m/s + 9.8 m/s)/2=4.8 m/s

(attained in 1 second

starting with the initial speed vi=0 m/s

& ending with the final speed vf=9.8 m/s

added together

& (then) divided by 2).

 

But Janus has shown us

that formula does NOT work

& that we need

double the mass (2*m=m+m)

(that extra kg, is)

for the opposite

& equal reaction

(of the air)

when m is the mass

of helicopter + (passengers + fuel + bagage, etc)

(Newton's 3rd law).

 

The correct formula

for a(n ideal) helicopter" (hover power)

using that method,

with the same average speed va=4.8 ms,

(& ignoring losses)

would be

P=2*F*va, or

P=2*m*g*va.

Edited by Capiert
0

Share this post


Link to post
Share on other sites

you are over complicating this, it is an energy calc

 

At sea level and at 15 °C air has a density of approximately 1.225 kg/m³ according to ISA.

 

Assuming your 1kg mass is your helicopter, you need to displace an equivalent mass of air with your helicopter blades. The amount of air displaced is the velocity of air through the blades x swept area ie the power developed is proportional to the cube of the wind speed. - losses in blades.

0

Share this post


Link to post
Share on other sites

My point is the factor "2".

 

I couldn't figure out

why the standard formula (without the 2)

did NOT work (til now).

 

"Double" the (helicopter) mass m

is needed

because of Newton's 3rd (law)

sets an opposite mass (=the air)

also in motion.

Edited by Capiert
0

Share this post


Link to post
Share on other sites

Janus' post demonstrates that a propulsion system is most efficient when the 'propellant' is moving at the same ( but negative ) velocity as the vehicle.

A simple example would be the efficiency of a turbofan engine over a turbojet at the subsonic speeds of an airliner.

Or the efficiency of a propeller ( or helicopter blades ) at even slower speeds.

 

And helicopters can hover in 'ground effect' ( approximate height equal to rotor diameter ) with much improved efficiency ( up to 10% ) because of pressure effects and reduced blade tip recirculation

0

Share this post


Link to post
Share on other sites

you are over complicating this, it is an energy calc

 

At sea level and at 15 °C air has a density of approximately 1.225 kg/m³ according to ISA.

 

Assuming your 1kg mass is your helicopter, you need to displace an equivalent mass of air with your helicopter blades. The amount of air displaced is the velocity of air through the blades x swept area ie the power developed is proportional to the cube of the wind speed. - losses in blades.

No.

It's a momentum calculation.

0

Share this post


Link to post
Share on other sites

My point is the factor "2".

I couldn't figure out

why the standard formula (without the 2)

did NOT work (til now).

"Double" the (helicopter) mass m

is needed

because of Newton's 3rd (law)

sets an opposite mass (=the air)

also in motion.

To get the math straight

I meant

the hover power (equation)

P=F*(v+2*vi)

has "twice" the initial speed vi

which is usually ignored

when assumed zero.

Where the speed difference

v=vf-vi

is the final speed vf

minus the initial speed vi.

 

The standard

power

P~F*v

is an approximation

(instead of exact(ly including initial speed vi, too)).

Edited by Capiert
0

Share this post


Link to post
Share on other sites

Is there a significant difference in the amount of force needed for levitation from the top of a very high place, ( e.g. Mount Everest ), compared with that needed at ground level?

0

Share this post


Link to post
Share on other sites

My first passing thought/guess would be that the force would be the same.... but the power required to generate that same force could be different due to the difference in air pressure. Obviously g would be slightly effected due to height, but I doubt the effect would be significant next to the air pressure difference.

0

Share this post


Link to post
Share on other sites

Janus' post demonstrates that a propulsion system is most efficient when the 'propellant' is moving at the same ( but negative ) velocity as the vehicle.

A simple example would be the efficiency of a turbofan engine over a turbojet at the subsonic speeds of an airliner.

Or the efficiency of a propeller ( or helicopter blades ) at even slower speeds.

 

And helicopters can hover in 'ground effect' ( approximate height equal to rotor diameter ) with much improved efficiency ( up to 10% ) because of pressure effects and reduced blade tip recirculation

From that

I can interpret (=assume)

for the -10 % power advantage

(the following equivalents:)

the initial_speed vi=-0.98 m/s

is negative

as a backwind (=air bounce_back, from the ground)

in the power equation

P=F*(v+2*vi),

where

the speed_difference v=8.82 m/s.

 

Or,

the average_speed va=4.41 m/s

for the equation

P=2*F*va

where the force F=Wt weight (=m*g)

of the helicopter (with pilot etc).

Edited by Capiert
0

Share this post


Link to post
Share on other sites

My first passing thought/guess would be that the force would be the same.... but the power required to generate that same force could be different due to the difference in air pressure. Obviously g would be slightly effected due to height, but I doubt the effect would be significant next to the air pressure difference.

 

Thanks,DrP. Maybe g and air-pressure at great height would even out any differences in power-to-weight ratios relevant to g and air-pressure at ground-level.

 

Capiert - can i ask, would air-bounce alter greatly in flight over different terrains such as rolling hills,deserts,or over the sea,where the ground/water is rough, very uneven or not solid, requiring in-flight corrections to the power needed to keep altitude steady at any height or speed? Are these factors taken into consideration when creating formulae for levitation?

Edited by Tub
0

Share this post


Link to post
Share on other sites

I have not (yet) estimated surface friction

from surface texture.

(That's a big deal.)

Thanks,Capiert. I appreciate that - I'd find it impossible.

0

Share this post


Link to post
Share on other sites

I suppose (=guess)

a hovercraft's apron

helps round_out (=generalize)

th(os)e surface differences, a bit,

(to an air_pressure cushion force),

decreasing that factor 2

closer to 1.

 

But the air_leak outward

at the bottom means

we could never have 1.

 

 

Hover in a (seated) tube:

 

I'd expect,

a fan('s force, balanced & aimed downward)

in a (sitting)

vertical tube

 

(with (the tube's) inner diameter

slightly larger than the fan blade's diameter,

& with the tube's bottom sealed (off)

by the ground (it's seated on)),

 

would have a better

merit factor,

 

& help get closer

to the 48 W/kg (instead of 96 W/kg)

due to the contained air's (almost strictly vertical (net))

bounce_back.

I.e. less air_leak sideways, as redirection of force (to 90 degrees=) horizontal.

 

The air_pressure

P=F/A

is obviously larger

under the fan

 

(for the (same) area A,

in any direction inside the tube),

 

to support the fan's weight.

Edited by Capiert
1

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!


Register a new account

Sign in

Already have an account? Sign in here.


Sign In Now
Sign in to follow this  
Followers 0