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Excess speed to height


mardlamock

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Hi everyone! So, I ve been thinking about how high a plane can climb given an initial airspeed and a minimum airspeed. I first thought it was a simple matter of potential+kinetic energy remaining equal throughout the flight, but then I brain farted and now I dont really know what to think.

If the only forces acting on the aircraft were gravity and drag, then the final mechanical energy would be always less than the initial, but that is never true because we also have lift. When you add lift into the equations, the rate at which you lose energy diminishes considerably, if you are ascending at a constant rate that is. This is because the vertical lift force cancels the weight of the aircraft, and the resulting energy losses are only due to drag and the horizontal lift force component.

Example: a plane is going at 200m/s, it performs a climb until it reaches 180m/s at a constant ascent rate of 5m/s. Without lift or drag the maximum height the plane coud reach would be 20m (1/2*20^2=g*h). Assuming that the angle with respect to the surface of the earth is small then the horizontal lift force component is negligible. In order for the final energy to be less than the initial energy, the acceleration due to drag has to be more than 5 m/s^2. But what if it isnt?

Lets say instead of it being 5m/s^2 its 4m/s^2, then, the plane will take 5 seconds to lose the 20m/s we were willing to convert into height. And as we said before, the climb rate remained constant at 5m/s, which means that the height will increase by of 25m. The thing is that, if it increased to 25 m then the system has more mechanical energy than at the start of the climb. So, what am i missing here? I have seriously got no idea where the extra energy comes from and wether its actually available for use.

Anyways, thanks for reading so far! Please reply and tell me what you think!

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If you have no lift or drag, then it's pure ballistic motion, and with a 5 m/s upward velocity, you'll reach an apex of about 1.27 m at about 0.5 s.

If you want a constant upward velocity you either need to be adding energy with an engine or you need to be converting your forward velocity into lift (and changing the rate at which you do so).

Edited by pincushionman
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Something else to consider is that a powered aircraft is not a constant-mass system. You're continually burning fuel decreasing the mass of the aircraft. If you're calculating using only potential and kinetic energy you cannot drop the mass term because the mass at say t = 5 seconds will not be the same as the mass at t = 6 seconds.

If you set up the equation so that you began at t1 and ended at t2 you might set it up something like this:

1/2*mt1*v12 + mt1*g*h1 = 1/2*mt2*v22 + mt2*g*h2

To find h2 we rearrange the equation:

{1/2*(mt1*v12 - mt2*v22) + mt1*g*h1}/mt2*g = h2

Doing a little simplification we get:

1/2*(mt1/mt2*v12 - v22)/g + mt1/mt2*h1 = h2

If we use the numbers from your example, and a couple of others:

v1 = 200 m/s

v2 = 180 m/s

We'll take:

h1 = 1000 m

mt1 = 5000 kg

mt2 = 4800 kg (we'll say that either this climb happens over a long period of time, or we have a really inefficient engine)

Thus we get:

h2 = 1/2*(5000/4800*2002 - 1802)/9.81 + 5000/4800*1000

h2 = 1/2*(1.042*40000 - 32400)/9.81 + 1.042*1000

h2 = 1/2*(9270)/9.81 + 1042

h2 = 1/2*945 + 1042

h2 = 1510 m

This is better than your example's 20m increase, but I think also your example calculation has a mistake. If you assume a constant mass then the mass terms do disappear entirely at which point you'd have:

If you take h1 = 0

1/2*(v12 - v22) = g*h2

1/2*(2002 - 1802) = g*h2

1/2*7600 = g*h2

Your equation appears to be:

1/2*(200 - 180)2 = g*h2

I hope I haven't been too vague in my equations, and that I've been able to help you gain at least a little better understanding. I believe that part of the confusion is arising from the fact that you can't consider the system to have a constant mass.

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Unless it is a plane powered by electric propellers. In this case there is no mass loss.

It depends on the power source. If the electricity is coming from solar panels, sure. If it's from an array of batteries or an internal combustion generator or an RTG or some other closed system power source, then it loses mass.

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I'm not sure why people are trying to bring in more complex factors when the calculations in the initial post are wrong.

If a plane is travelling at 200 m/s forward, and ascends to the point that it reaches 180 m/s, then regardless of the rate it does so at or its aerodynamic efficiency, the absolute maximum height it can gain is:

E1 = 1/2 200 ^2 = 20000 J/kg

E2 = 1/2 180 ^2 = 16200 J/kg

E1,2 = 20000 - 16200 = 3800 J/kg

h = E/mg = 3800 J / 9.81 m/s^2 = 387.36 m

Which is drastically higher than initially calculated. Of course, it'd never be able to reach that height, as drag would steal energy on the way, and it'd have to follow a ballistic ascent, meaning non-linear vertical speed, but that's the maximum possible without engines adding in energy. If you're considering engines, then the arguments about energy exchange go out the window since you're able to constantly add more. As for your calculations with the vertical acceleration, they're even more wrong, as you're starting with a plane that is travelling 200 m/s horizontally (or mostly horizontally), but then trying to determine accelerations from its vertical motion, which is unknown without quite a few further calculations.

Addendum: it's also kind-of a wrong question. If you'd started with the correct values, then you might've come up with a particular minimum deceleration the aircraft experiences. You then ask what happens if it accelerates less than that, and surprise, the numbers that come out say the aircraft has managed to gain energy - well, of course! That's why that's the minimum deceleration, because a lower deceleration breaks physics, and is therefore impossible. It'd be like if I gave you a baseball and asked you to, using only your own arm and no extensions, throw it a kilometre up - there simply isn't the energy available in human muscles to come close to that. You can calculate what it'd require - 1.46 kJ of energy and a starting velocity of 198.1 m/s - but that doesn't mean the system we're considering contains the capability to achieve that.

Edited by Iskierka
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It depends on the power source. If the electricity is coming from solar panels, sure. If it's from an array of batteries or an internal combustion generator or an RTG or some other closed system power source, then it loses mass.

Ehm... Do batteries really lose mass when delivering their charge ? I'd like to know how ! I guess it would be insignificant anyway.

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Ehm... Do batteries really lose mass when delivering their charge ? I'd like to know how ! I guess it would be insignificant anyway.

Yes, they do. Electrons are matter, and matter has mass. It is going to be insignificant, but it does lose mass. :)

Edited by SuperFastJellyfish
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Batteries don't typically lose electrons, they just move them. You'll get some mass loss from the lowered energy content, buy that'll be even more insignificant.

Yeah that's what i was thinking. Electrons move inside the circuits, but do not leave the system.

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