Workout from Flying

[McC]

One thing I've always kind of wondered about is how much energy it'd take for a human to fly about under his own power. Obviously, we're delving into the fantastic by even considering this, but set that aside a moment. Imagine a human being is capable of flight without a wing-like or exhaust-based mechanism (essentially, flight like a superhero character). How much energy would it take to stay aloft, accelerate, and so forth? Is it comprable to walking? Running? Sprinting?

My initial assumption was to suggest that the energy requied to levitate to a certain height could be surmised from PE = mgh. From there, the energy to fly about would be Fd (essentially, only applying energy to accelerate or decelerate). This math feels vaguely wrong, but I'm not really sure.

Anyway, my initial calculations (which I feel are incomplete and inaccurate) are based on the following initial parameters:

Code: Select all

Mass of flyer:           180 lbs. /  81.7 kg
Altitude of flight:       40 ft.  /  12.2 m)
Cruising speed:           80 mph  / 129 kph
Cruise Acceleration Time: 10 seconds

So, the questions:
How much energy/power is required to fly up to an altitude of 12.2 m and maintain it?
How much energy/power is required to accelerate to 129 kph in 10 seconds?
Corrolary to the above: How much energy is required to maintain speed after acceleration ceases while in an atmosphere, accounting for drag? This question is less important to me, since it requires some heavy-hitting math, as I understand it.

Thanks.

[Feil]

Disregarding for a moment that energy/power is time... this is a basic first-semester physics problem. It tells me that you should take first semester physics. Everybody should take first-semester physics...



Feil approaches the blackboard

Ahem.

Part 1: energy to attain altitude

Delta Energy = mgh, where m is the mass of Superman, g is acceleration due to gravity (9.8m/s^2) and h is the difference Superman's final altitude (12.2m) and his original altitude (0).

Power = mgh/t, where t is the time it took Superman to get that high (unspecified in OP).

Part 2: energy to attain cruising velocity

There will be a bit of air resistance for the first ten seconds, but since the time interval is so slight relative to the amount of energy being used, there's not much point in worying about it. Hence,

Assume no air resistance.

Delta Energy = .5mv^2 where v is Superman's final velocity.

Power = (.5mv^2)/t, t being the time it took Superman to attain that velocity.

Part 3: energy to maintain cruising speed

Assume Superman is a cylendar with a base of radius .3m (area=.3m^2).

Further assume that there is no turbulence-created drag on Superman (IE that the only factors are cross-sectional area and velocity).

Force=ma, and Superman will be displacing X=(1.2kg/m^3)vA of air, where v is his velocity and A his cross-sectional area, value expressed in kg/s.

This amount of kilograms [disregard seconds, we'll pick them back up in a second] must be displaced an average of Y=1-sqrt(2)*r, value expressed in meters, in one second , for an average velocity of Y/s, assume linear acceleration, so final velocity of 2Y/s, meaning an acceleration of 2Y/s^2.

Therefore, the amount of force Superman is exerting is 2XY Newtons.

Power is force*velocity, so Superman must provide a Power of 2XYv watts.

Energy is P*t, t being the durration of Superman's flight.

Part 4: energy to maintain altitude
Because the only force Superman is being acted on in the direction of h is that of gravity, and because h is not changing, F*v=F*0=0, ergo there is no energy or power required to keep Superman aloft.

Superman must supply a force of mg to keep from falling, but this requires no more energy than your chair does to keep you from falling.

[McC]

Disregarding for a moment that energy/power is time... this is a basic first-semester physics problem. It tells me that you should take first semester physics. Everybody should take first-semester physics...

I took both Honors and AP (college) physics in high school. It's just been a while. It's less that I don't know how, and more than I'm uncertain.

Part 1: energy to attain altitude

Delta Energy = mgh, where m is the mass of Superman, g is acceleration due to gravity (9.8m/s^2) and h is the difference Superman's final altitude (12.2m) and his original altitude (0).

Power = mgh/t, where t is the time it took Superman to get that high (unspecified in OP).

This is how I approached it when I first tried to work it out, and it comes out to about 9.8 kilojoules.

Part 2: energy to attain cruising velocity

There will be a bit of air resistance for the first ten seconds, but since the time interval is so slight relative to the amount of energy being used, there's not much point in worying about it. Hence,

Assume no air resistance.

Delta Energy = .5mv^2 where v is Superman's final velocity.

Power = (.5mv^2)/t, t being the time it took Superman to attain that velocity.

Using the KE equation, the energy in question is 52.4 kilojoules.

When I approached it, I approached it from E=Fd. F, of course, being comprised of m (81.7 kg) and a, (dv/dt or 3.58 m/s^2), and d being defined by d = d(0)t + 1/2at^2 or 179 m. From that, you get the same value.

So, I obviously took the long way around.

Part 3: energy to maintain cruising speed

Assume Superman is a cylendar with a base of radius .3m (area=.3m^2).

Further assume that there is no turbulence-created drag on Superman (IE that the only factors are cross-sectional area and velocity).

Force=ma, and Superman will be displacing X=(1.2kg/m^3)vA of air, where v is his velocity and A his cross-sectional area, value expressed in kg/s.

This amount of kilograms [disregard seconds, we'll pick them back up in a second] must be displaced an average of Y=1-sqrt(2)*r, value expressed in meters, in one second , for an average velocity of Y/s, assume linear acceleration, so final velocity of 2Y/s, meaning an acceleration of 2Y/s^2.

Therefore, the amount of force Superman is exerting is 2XY Newtons.

Power is force*velocity, so Superman must provide a Power of 2XYv watts.

Energy is P*t, t being the durration of Superman's flight.

This is the part of the equation I mostly disregarded for the sake of simplicity (i.e. doing this flight in a vacuum). I'll have to come back and look at it more closely to get a precise value.

Part 4: energy to maintain altitude
Because the only force Superman is being acted on in the direction of h is that of gravity, and because h is not changing, F*v=F*0=0, ergo there is no energy or power required to keep Superman aloft.

Superman must supply a force of mg to keep from falling, but this requires no more energy than your chair does to keep you from falling.

This is the part that bothers me. If someone levitates to a given height with nothing supporting them other than their levitation (which ostensibly consumes energy to maintain), if they stop attempting to project that levitation, gravity will take over and they will fall. So to suggest that someone can be 'placed' at a given height with no support beneath them and will not fall is absurd. However, I understand what you're saying with regard to the whole 'no work is being done', since work (energy) necessitates a distance through which a force acts. How, then, to answer this question?

Would it be wrong to say that they must expend the energy defined by mgh each second to maintain their height (neglecting things like lift and such that they may or may not gain some additional help from in the course of moving), lest gravity resume its hold on them and pull them back down?

[Feil]

This is the part that bothers me. If someone levitates to a given height with nothing supporting them other than their levitation (which ostensibly consumes energy to maintain), if they stop attempting to project that levitation, gravity will take over and they will fall. So to suggest that someone can be 'placed' at a given height with no support beneath them and will not fall is absurd. However, I understand what you're saying with regard to the whole 'no work is being done', since work (energy) necessitates a distance through which a force acts. How, then, to answer this question?

Would it be wrong to say that they must expend the energy defined by mgh each second to maintain their height (neglecting things like lift and such that they may or may not gain some additional help from in the course of moving), lest gravity resume its hold on them and pull them back down?

Yes, it would be wrong. It takes absolutely no energy to maintain altitude; if this were not so your posterior would be churning out power by the kilowatt to keep you from falling through your chair straight to the center of the Earth.

What it does require is force. That force can be supplied by lift (which is just another name for pressure), by another gravitational body, by a large-scale electromagnetic force field, or by the microscopic force-fields holding the atoms of your chair together.

You can use energy to create a force, by accelerating something in the downward direction (like air, in the case of a helecoptor, or the ejecta from your rocket), which is what you'll have to do if you don't have a magical force-creator-thing like Superman has.

However, if you generate energy in the process of generating a force without accelerating something, then that energy has to go somewhere, and I don't recall seeing Superman glow...

[Howedar]

Part 3: energy to maintain cruising speed

Assume Superman is a cylendar with a base of radius .3m (area=.3m^2).

Further assume that there is no turbulence-created drag on Superman (IE that the only factors are cross-sectional area and velocity).

Force=ma, and Superman will be displacing X=(1.2kg/m^3)vA of air, where v is his velocity and A his cross-sectional area, value expressed in kg/s.

This amount of kilograms [disregard seconds, we'll pick them back up in a second] must be displaced an average of Y=1-sqrt(2)*r, value expressed in meters, in one second , for an average velocity of Y/s, assume linear acceleration, so final velocity of 2Y/s, meaning an acceleration of 2Y/s^2.

Therefore, the amount of force Superman is exerting is 2XY Newtons.

Power is force*velocity, so Superman must provide a Power of 2XYv watts.

Energy is P*t, t being the durration of Superman's flight.

This is an interesting approach; unfortunately it's wrong. Viscosity is necessary to have drag. When you neglect turbulence (by which I think you probably mean viscous effects in general), there can be no drag.

This paradox was pretty frustrating in the 1700s; a guy named D'Alembert had some real issues with it.

[McC]

Yes, it would be wrong. It takes absolutely no energy to maintain altitude; if this were not so your posterior would be churning out power by the kilowatt to keep you from falling through your chair straight to the center of the Earth.

What it does require is force. That force can be supplied by lift (which is just another name for pressure), by another gravitational body, by a large-scale electromagnetic force field, or by the microscopic force-fields holding the atoms of your chair together.

You can use energy to create a force, by accelerating something in the downward direction (like air, in the case of a helecoptor, or the ejecta from your rocket), which is what you'll have to do if you don't have a magical force-creator-thing like Superman has.

However, if you generate energy in the process of generating a force without accelerating something, then that energy has to go somewhere, and I don't recall seeing Superman glow...

Ok, I see where the disconnect is happening -- I'm not phrasing things correctly. I'm thinking of energy in terms of providing "juice" to generate the force, not in terms of actual work being done (force applied through distance).

So, phrasing the original question more appropriately, how much "bodily fuel" would generating/applying these various forces necessitate?

Of course, it may be utterly impossible to answer that question, since answering it would necessitate knowing what kind of fuel efficiency this force-generation effect has...

[Feil]

Of course it's wrong; it's a vastly oversimplified model. Neither is Superman a cylinder.

However, if my guestimation of the cross-sectional area of Superman is approximately correct, my model should give us a fairly close low-end estimate, which is really all we need. The effects of viscosity on Superman (and we would need to use a much more complex Superman model) would be very difficult to calculate, but, fortunately, they are insignificant relative to the requirement just to get the air out of the way, so we can ignore them.

[Howedar]

Of course it's wrong; it's a vastly oversimplified model. Neither is Superman a cylinder.

However, if my guestimation of the cross-sectional area of Superman is approximately correct, my model should give us a fairly close low-end estimate, which is really all we need. The effects of viscosity on Superman (and we would need to use a much more complex Superman model) would be very difficult to calculate, but, fortunately, they are insignificant relative to the requirement just to get the air out of the way, so we can ignore them.

I will repeat myself. Without viscosity, and assuming solidly subsonic speeds, drag in all cases is zero.

[Steel]

I will repeat myself. Without viscosity, and assuming solidly subsonic speeds, drag in all cases is zero.

I think that the general rough formula for drag is:

F= Cd*R*A*v^2

Where we have R being the density of the medium,
A cross sectional area of body
v speed of body
and Cd a drag coefficient for the body depending on exact profile (aka its a fudge factor) and is dimensionless

And this is even dimensionally correct!

So all Feil has done is to assume that the Cd is about 1, essentially

[Steel]

ah bugger... unfiortunately i dont think that speed is low enough here to approximate v=v^2

But if you use the above formula with density of air = 1kg/m^3 and take Cd as 1 it might give a reasonably accurate answer.

To get energy to accelerate to V need to integrate F between s=0 and s=S

[Darth Wong]

One thing I've always kind of wondered about is how much energy it'd take for a human to fly about under his own power. Obviously, we're delving into the fantastic by even considering this, but set that aside a moment. Imagine a human being is capable of flight without a wing-like or exhaust-based mechanism (essentially, flight like a superhero character). How much energy would it take to stay aloft, accelerate, and so forth? Is it comprable to walking? Running? Sprinting?

Without an actual mechanism of flight, this question is totally unanswerable. Simply quantifying the drag forces and KE/GPE of any given flight vector does not even remotely answer the question, and certainly doesn't allow fair comparison with walking, running, or sprinting, all of which convert large amounts of energy into forms other than forward movement.

[raptor3x]

The first two are pretty much trivial so I'll only answer no. 3. If we assume a best case where you have a propulsive efficiency of 1 and you don't have any kind of induced to drag to deal with then we can get a low end answer to show how implausible it is for humans to fly, at least with the airspeed you specified.

The first thing we need to get is is some way to predict how much drag force will be exerted at any given velocity. This can be expressed as such:

D = qf (1)

where q is the dynamic pressure, (1/2)rho*V^2, and f is the equivalent flat plate area, or C_dA (which is the drag coefficent times the frontal area.) I got an approximation of f from the terminal velocity of a sky diver in a vertical diving position. No altitude was given for that number so I assumed sea level conditions, which is most likely wrong but it doesn't matter much. Using a weight of 160 lbs, the terminal velocity of 80 m/s, and 1.29 kg/m^3 for density produces an f of 0.172 m^2. Now we can get a power required to maintain level flight.

Using equation (1) and plugging in our values of f and the dynamic pressure at 80 mph, we get a power required of 5.1 kW. Now to put this into perspective, Lance Armstrong on his best climb up Le Alp De 'Huez averaged somewhere around 500 Watts for around 30 minutes. Also, these fellows can generate up around 2 kW for about 10-12 seconds.

Track Pros

Needless to say, it's not going to happen.