A Fascinating Mathematical Proof on Missile Kinetic Energy
Nov 10, 2016 3:34:16 GMT
apophys, Durandal, and 5 more like this
Post by Crowne on Nov 10, 2016 3:34:16 GMT
So here's an interesting question.
Say you have a missile. You want to accelerate this missile to have the maximum Kinetic Energy possible, which means having the maximum mass and velocity possible, since K = 0.5 * m * v2.
However, as you burn through your Delta-V you lose some of your mass, which costs you Kinetic Energy, but you gain velocity, which seems more important since it's squared in the function.
Nevertheless, as you perform the experiment and expend all your fuel to achieve your maximum velocity, you notice your kinetic energy actually starts to drop past a certain point!
Indeed, there did exist a point after which it was useless to accelerate further, since the net change in Kinetic Energy will be negative.
Mathematically, what happened? Is this always the case?
The answer to the latter question is no. This point of maximum Kinetic Energy beyond which you suffer a net loss only exists if your wet mass to dry mass ratio is greater than Euler's constant, squared. (e2)
What about momentum? Does it behave similarly?
The answer is yes, a similar point exists for momentum, but only if your wet mass to dry mass ratio is greater than Euler's constant. (e)
Therefore, if your ratio is less than e, you will not run into this problem.
To prove these conjectures of mine, and for your mathematical enjoyment, here is a modifiable graph that demonstrates it all:
www.desmos.com/calculator/ect9th2wao
The black vertical lines are the domain of the function during which it is valid. Anything beyond them is physically impossible since after them the rocket begins to have "negative" fuel with "negative" mass.
The axis x is time.
The red K(x) function is the Kinetic Energy.
The dashed green m(x) function is the total mass of the rocket at a given point in time.
The (hidden) dashed green F(x) function is the remaining fuel of the rocket.
The dashed purple v(x) function is the velocity of the rocket.
The (hidden) dashed purple RemainingDeltaV(x) function is, clearly, the function that describes the remaining Delta-V of the craft.
The solid blue p(x) function is the momentum of the rocket.
The wet mass to dry mass ratio is calculated at the end. In the example given, it is 11.
However, if you wish to modify it to see how my earlier observations about the maxima existing only if the ratios are e or e^2, see this second interactive graph, and set the ratio to those values.
Cheers,
Crowne
Say you have a missile. You want to accelerate this missile to have the maximum Kinetic Energy possible, which means having the maximum mass and velocity possible, since K = 0.5 * m * v2.
However, as you burn through your Delta-V you lose some of your mass, which costs you Kinetic Energy, but you gain velocity, which seems more important since it's squared in the function.
Nevertheless, as you perform the experiment and expend all your fuel to achieve your maximum velocity, you notice your kinetic energy actually starts to drop past a certain point!
Indeed, there did exist a point after which it was useless to accelerate further, since the net change in Kinetic Energy will be negative.
Mathematically, what happened? Is this always the case?
The answer to the latter question is no. This point of maximum Kinetic Energy beyond which you suffer a net loss only exists if your wet mass to dry mass ratio is greater than Euler's constant, squared. (e2)
What about momentum? Does it behave similarly?
The answer is yes, a similar point exists for momentum, but only if your wet mass to dry mass ratio is greater than Euler's constant. (e)
Therefore, if your ratio is less than e, you will not run into this problem.
To prove these conjectures of mine, and for your mathematical enjoyment, here is a modifiable graph that demonstrates it all:
www.desmos.com/calculator/ect9th2wao
The black vertical lines are the domain of the function during which it is valid. Anything beyond them is physically impossible since after them the rocket begins to have "negative" fuel with "negative" mass.
The axis x is time.
The red K(x) function is the Kinetic Energy.
The dashed green m(x) function is the total mass of the rocket at a given point in time.
The (hidden) dashed green F(x) function is the remaining fuel of the rocket.
The dashed purple v(x) function is the velocity of the rocket.
The (hidden) dashed purple RemainingDeltaV(x) function is, clearly, the function that describes the remaining Delta-V of the craft.
The solid blue p(x) function is the momentum of the rocket.
The wet mass to dry mass ratio is calculated at the end. In the example given, it is 11.
However, if you wish to modify it to see how my earlier observations about the maxima existing only if the ratios are e or e^2, see this second interactive graph, and set the ratio to those values.
Cheers,
Crowne
