How much Power do I need to levitate 1 kg?

[Capiert]

1 kg has a weight force ~9.8 Newton.

What is the power relation (to balance force)?

 

E.g. a helicopter doesn't stay up in the air for nothing,

but (instead) burns chemical energy

at a specific rate.

I'm NOT thinking (=talking) about Zeppelins (e.g. buoyancy: force vs force balance).

Edited April 26, 2017 by Capiert

[DrKrettin]

It depends a lot on the efficiency of what you are using to lift it.

[swansont]

If you levitate it with light reflection, it's about 1.21 GW

 

http://blogs.scienceforums.net/swansont/archives/60

[Sriman Dutta]

power = force x velocity

[Capiert]

It depends a lot on the efficiency of what you are using to lift it.

 

I'm assuming no frictional losses,

so I guess 100% for the ideal case.

If you levitate it with light reflection, it's about 1.21 GW

 

http://blogs.scienceforums.net/swansont/archives/60

Gasp! That's too expensive, for my fuel bill.

Is there another way to do it?

power = force x velocity

That's the problem:

when hovering the helicopter does not move:

e.g. speed (velocity) is zero. ?

Edited April 26, 2017 by Capiert

[swansont]

 

That's the problem:

when hovering the helicopter does not move:

e.g. speed (velocity) is zero. ?

 

 

The CoM is not moving, but the air is. You can deduce the power from that.

[Capiert]

I'm sorry, I can't quite follow yet, M=Mass?

Do you mean 1 kg of air is moving, per second?

[Strange]

Gasp! That's too expensive, for my fuel bill.

Is there another way to do it?

 

 

Rope?

[Capiert]

Nope, it doesn't burn rope.

 

(Old ropes were made of hemp (=grass),

but burning them to smoke

won't make you high. )

 

Please continue though.

Edited April 26, 2017 by Capiert

[Sensei]

 

The CoM is not moving, but the air is. You can deduce the power from that.

 

I'm sorry, I can't quite follow yet, M=Mass?

Do you mean 1 kg of air is moving, per second?

 

CoM is shortcut from Center of Mass

https://en.wikipedia.org/wiki/Center_of_mass

If you levitate it with light reflection, it's about 1.21 GW

 

http://blogs.scienceforums.net/swansont/archives/60

 

But where are equations leading to this value, to follow by people step by step.. ?

That's the problem:

when hovering the helicopter does not move:

e.g. speed (velocity) is zero. ?

 

Helicopter is moving air from the above of rotor to below.

Do you have electric drone? Run it and you should feel stream of air below it.

Edited April 26, 2017 by Sensei

[Capiert]

I don't have an electric drone,

but an electric fan.

 

I'm still lost as to how

you all would calculate

the P vs F relation.

 

Would someone please continue

& do the calculations?

[swansont]

 

But where are equations leading to this value, to follow by people step by step.. ?

 

 

Photon momentum is E/c. If we look at this per unit time, this will be E/ct, which is the force. E/t is power. Thus the force is P/c. There's a factor of 2 because of reflection, and is assumed to be perfect. So the total force of reflection is 2P/c

 

That gets you a little more than 8N of force. 1 kg requires 9.8N

I don't have an electric drone,

but an electric fan.

 

I'm still lost as to how

you all would calculate

the P vs F relation.

 

Would someone please continue

& do the calculations?

 

How fast does the air move as it goes through the fan, and what is the fan's area?

 

https://www.grc.nasa.gov/www/k-12/VirtualAero/BottleRocket/airplane/thrsteq.html

https://www.wired.com/2012/06/how-hard-is-the-human-powered-helicopter/

[John Cuthber]

If you levitate it with light reflection, it's about 1.21 GW

 

http://blogs.scienceforums.net/swansont/archives/60

Or with precisely zero Watts if you use a shelf.

So the answer seems to be "anywhere between none and lots- depending how you do it"

[imatfaal]

In the spirit of the OP - and in that I am assuming that levitate means hold in the air without an immediately obvious means of support - I suppose one could make the kilo weight into a sphere and support it in an airflow like the pingpong ball at the end of a straw (which is used in demonstrations of Bernoulli).

[Capiert]

Photon momentum is E/c. If we look at this per unit time, this will be E/ct, which is the force. E/t is power. Thus the force is P/c. There's a factor of 2 because of reflection, and is assumed to be perfect. So the total force of reflection is 2P/c

 

That gets you a little more than 8N of force. 1 kg requires 9.8N

 

 

How fast does the air move as it goes through the fan, and what is the fan's area?

 

https://www.grc.nasa.gov/www/k-12/VirtualAero/BottleRocket/airplane/thrsteq.html

https://www.wired.com/2012/06/how-hard-is-the-human-powered-helicopter/

The 2nd link seems to be in the direction I want (it to go).

But as John seems to indicate,

it looks like (some degree of) guesswork.

I thought physics was (suppose to be) an exact science.

I can't believe this is an impossible task.

To keep things clear

I'm looking for an electrical equivalent (equation)

but not done with light.

I thought it would have been easy for you all

to predict helicopter fuel costs

(& I still have faith in you can).

In the spirit of the OP - and in that I am assuming that levitate means hold in the air without an immediately obvious means of support - I suppose one could make the kilo weight into a sphere and support it in an airflow like the pingpong ball at the end of a straw (which is used in demonstrations of Bernoulli).

Yes, that's doing it backwards though.

I'm interested in hovering a helicopter.

Fuel costs.

 

I suppose things get more interesting

when we bring the hovering height h (e.g. 1 m)

into the picture.

E.g. It takes more power to hover higher (e.g. 2, 10, or even 100 or 1000 m);

even thought that is further away from

the earth's center (of mass).

Your gravity laws say

the force should decrease instead

(at higher heights).

Edited April 26, 2017 by Capiert

[Mordred]

Its a relatively exact science if were provided the details we require instead of guessing.

 

Can yoy build something without knowing what parts you have to work with?

 

Can you engineer a bridge without knowing how much stress the metal can take?

 

Give us some specifications on the equipment you have to work with and remove the guess work. Then maybe just maybe we can give you the accurate answers your seeking.

 

How big a fan, how many blades, whst is the blade length,what is the blade pitch? what efficiency is your motor,

 

details are needed if you have any datasheets, part numbers or specifications sheets those are helpful as well

Edited April 27, 2017 by Mordred

[DrKrettin]

I suppose things get more interesting

when we bring the hovering height h (e.g. 1 m)

into the picture.

E.g. It takes more power to hover higher (e.g. 2, 10, or even 100 or 1000 m);

even thought that is further away from

the earth's center (of mass).

Your gravity laws say

the force should decrease instead

(at higher heights).

 

Assuming that the heights are such that the change in g is negligible, perhaps you can explain why it takes more power to hover higher. It takes more work to get it there, but that's not the issue.

[swansont]

 

Assuming that the heights are such that the change in g is negligible, perhaps you can explain why it takes more power to hover higher. It takes more work to get it there, but that's not the issue.

 

 

Lift/thrust depends on the mass flow rate of the air. Less air, less flow, less thrust. So you have to crank up the rotor speed (or something equivalent) to get the same thrust.

[DrKrettin]

 

 

Lift/thrust depends on the mass flow rate of the air. Less air, less flow, less thrust. So you have to crank up the rotor speed (or something equivalent) to get the same thrust.

 

Yes, agreed, but the it's not clear to me that this would require more power, and he's assuming 100% efficiency. Ignoring buoyancy, is it the case that the denser the atmosphere, the less the power needed to hover?

[swansont]

Yes, agreed, but the it's not clear to me that this would require more power, and he's assuming 100% efficiency. Ignoring buoyancy, is it the case that the denser the atmosphere, the less the power needed to hover?

He's assuming no frictional losses, but that's not all that's included in efficiency, and it seemed that you were asking a question independent of that assumption, i.e. in the real world.

 

In any event, the air resistance of the blade depends on the speed^2, which we've increased. I suspect that is a contributor to the increased power.

[Handy andy]

1 kg has a weight force ~9.8 Newton.

What is the power relation (to balance force)?

 

E.g. a helicopter doesn't stay up in the air for nothing,

but (instead) burns chemical energy

at a specific rate.

I'm NOT thinking (=talking) about Zeppelins (e.g. buoyancy: force vs force balance).

 

Its an energy calculation.

 

A helicopter blade sweeps an area given by 3.14r²​ the speed of the air through the blade is m/s. The overall power is therefore proportional to cube of the wind speed going through the blades​ . They will have losses

 

Power = wT where T = Torque, and w the speed of the helicopter blade. Power = J/s you are trying to overcome the gravitational energy on the mass, so to hold it in position ..... is this a homework question.

 

What mass of air do you need to displace to hold 1 kg mass in the air. What is the density of air, how much do you need to move to act against gravity.

Edited April 27, 2017 by Handy andy

[Janus]

 

Yes, agreed, but the it's not clear to me that this would require more power, and he's assuming 100% efficiency. Ignoring buoyancy, is it the case that the denser the atmosphere, the less the power needed to hover?

Let's look at it this way: The upwards force you need to apply to a 1kg mass in order to support it against gravity is the same as it would take to accelerate the mass at 9.8m/s² with no gravity.

So let's assume that you are going to provide this by throwing a mass backwards and benefiting from the reaction force. (which is basically what a helicopter does)

We'll start with the assumption that you are throwing a 1 kg mass backwards, and this takes 1 sec to get it up to speed. After the one sec, the 1kg mass you threw is moving at 9.8 m/s and your 1kg mass is moving at 9.8m/s per sec in the other direction. the total energy your need to expend will be 2*1kg* (9.8m/s)²/2 = 96.04 Joules. Over one sec that is a power of 96.04 watts

 

Now imagine that you are throwing a 100g mass backwards to provide the same acceleration for your 1 kg mass over 1 sec. In order for the momentum of the 100g mass mass to balance the momentum of your mass once your mass reaches 9.8m/sec, the 100g will have to be moving at 98m/sec. The total final energy of the system and thus the energy you would have to expend is now 1kg*(9.8m/s)²/2 + .1 kg*(98m/s)²/2 = 528.22 joules. Over 1 sec that is a power of 528.22 watts. Your power usage rate went up by a factor of 5.5.

[DrKrettin]

Let's look at it this way: The upwards force you need to apply to a 1kg mass in order to support it against gravity is the same as it would take to accelerate the mass at 9.8m/s² with no gravity.

So let's assume that you are going to provide this by throwing a mass backwards and benefiting from the reaction force. (which is basically what a helicopter does)

We'll start with the assumption that you are throwing a 1 kg mass backwards, and this takes 1 sec to get it up to speed. After the one sec, the 1kg mass you threw is moving at 9.8 m/s and your 1kg mass is moving at 9.8m/s per sec in the other direction. the total energy your need to expend will be 2*1kg* (9.8m/s)²/2 = 96.04 Joules. Over one sec that is a power of 96.04 watts

 

Now imagine that you are throwing a 100g mass backwards to provide the same acceleration for your 1 kg mass over 1 sec. In order for the momentum of the 100g mass mass to balance the momentum of your mass once your mass reaches 9.8m/sec, the 100g will have to be moving at 98m/sec. The total final energy of the system and thus the energy you would have to expend is now 1kg*(9.8m/s)²/2 + .1 kg*(98m/s)²/2 = 528.22 joules. Over 1 sec that is a power of 528.22 watts. Your power usage rate went up by a factor of 5.5.

 

That is illuminating, thanks. So the energy expended is a function of the square of the speed of the mass thrown backwards, and the momentum a direct ratio. Thus the greater the mass used, the less the power usage. So presumably, the denser the medium in which you are working, the easier it is to hover.

[John Cuthber]

But as John seems to indicate,

it looks like (some degree of) guesswork.

I thought physics was (suppose to be) an exact science.

I can't believe this is an impossible task.

 

That's a bit like asking "how big is a car?", then complaining when someone says that it depends- it could be this

https://en.wikipedia.org/wiki/Peel_P50

or this

https://en.wikipedia.org/wiki/Ford_Super_Duty#Third_generation_.282011.E2.80.93present.29

[Capiert]

I think Janus did a marvelous job

(that's a plus point for women('s intuition), if so,

(over men's details, although both are needed as a team)).

I would have tipped ~48 W/kg to levitate, (#7),

but as Janus has shown us

a helicopter needs double that to hover,

(because 2 masses are moved: helicopter & air (also))

so there is at least -50% loss with a helicopter.

 

Janus('s method or track)

has solved the question (so far)

the way I intended

(simply, (& exactly) with com vs coe, conservation of momentum vs energy)

as a physics feat (overview),

(instead of (guessing, back) engineering ((unavailable) details), yet;

which would be the next step).

 

Thanks Janus.

 

Your 2nd example (0.1 kg)

disturbingly indicates

to me a(n inferiority)

problem of energy

which Swansont denies

(in my Mechanical Symmetry thread #1 (theme), #12 (denial), #77 (request))

as non_existent.

Edited May 1, 2017 by Capiert