- 18 Jan '10 01:37So you have a field accelerator that you can use on a spacecraft that gives you the same efficiency at 100% of a car on a road. In space, to get say, 1 g of accel, you need like gigawatts because of the terrible efficiency rating of heating stuff to shove it out the rear end in a rocket.

So you have a magic anti-matter power supply generating 1 megawatt pretty much forever. So with that amount of power converted 100% into kinetic energy to accelerate a spacecraft, how heavy a spacecraft can it push with one G of acceleration? - 18 Jan '10 04:13 / 3 edits

Harder than it seems!*Originally posted by sonhouse***So you have a field accelerator that you can use on a spacecraft that gives you the same efficiency at 100% of a car on a road. In space, to get say, 1 g of accel, you need like gigawatts because of the terrible efficiency rating of heating stuff to shove it out the rear end in a rocket.**

So you have a magic anti-matter power supply generating 1 megawatt p ...[text shortened]... nergy to accelerate a spacecraft, how heavy a spacecraft can it push with one G of acceleration?

W/t = P

Fd/t = P

Fv = P

v = P/F

So from power and force we can find velocity (?). Does this mean the power needed for acceleration varies with velocity?

Edit - Power and force gives momentum. Power and acceleration (what we have here) gives velocity. - 18 Jan '10 04:22

P = Fv*Originally posted by AThousandYoung***Harder than it seems!**

W/t = P

Fd/t = P

Fv = P

v = P/F

So from power and force we can find velocity (?). Does this mean the power needed for acceleration varies with velocity?

P = Mav

P/av = M

We need to know the velocity for some reason to solve this I think, though I don't understand why. - 18 Jan '10 05:23 / 2 edits

I know, but that's how the equations work out.*Originally posted by sonhouse***What velocity? You are starting out at zero!**

Power = Force x Velocity

We have Power; we have acceleration, and mass is unknown, so Force is unknown; but then there's that velocity term. WTF does it mean?

Maybe it's change in velocity; delta v. - 18 Jan '10 05:27

So perhaps it can be interpreted as:*Originally posted by AThousandYoung***I know, but that's how the equations work out.**

Power = Force x Velocity

We have Power; we have acceleration, and mass is unknown, so Force is unknown; but then there's that velocity term. WTF does it mean?

Maybe it's change in velocity; delta v.

"Power is equal to the mass times the acceleration times the amount that the object actually changes velocity."

It doesn't make sense... - 18 Jan '10 06:24

Well given enough time, the object would accelerate such that v-->c, during which its mass is constantly changing, m--> infinity. So it would be dependent on velocity. But then again your not using differential equations....just a thought.*Originally posted by AThousandYoung***So perhaps it can be interpreted as:**

"Power is equal to the mass times the acceleration times the amount that the object actually changes velocity."

It doesn't make sense... - 18 Jan '10 08:59

Yeah, that crossed my mind too, but I thought it unlikely that that was the cause of the v in that equation.*Originally posted by joe shmo***Well given enough time, the object would accelerate such that v-->c, during which its mass is constantly changing, m--> infinity. So it would be dependent on velocity. But then again your not using differential equations....just a thought.** - 18 Jan '10 16:21 / 8 edits

Well here is a hint: Suppose you had a rail car, like a dragster. The track would be a mile long gear basically, with metal grooves perpendicular to the track, got that part?*Originally posted by AThousandYoung***Yeah, that crossed my mind too, but I thought it unlikely that that was the cause of the v in that equation.**

Then the wheels of the car would match the grooves and the friction would be close to zero because they are coated with super teflon.

The front and back of the car would have bearings in a groove going down the center of the track, keeping the car perfectly straight and level the whole run.

Now you have an antimatter power source, weighs only a few pounds so it doesn't effect the problem and it produces 1 megawatt of energy.

That is coupled to a room temperature superconducting electric motor coupled at 100% efficiency to the wheels. So you hit the big power switch, putting the pedal to the metal so to speak.

So now how much can the total car mass and still give one G of acceleration?

We are not talking relativistic effects here, strictly say, below half of c. - 18 Jan '10 17:17

Well however you solved it, I would like to see...can you PM me the solution?*Originally posted by sonhouse***Well here is a hint: Suppose you had a rail car, like a dragster. The track would be a mile long gear basically, with metal grooves perpendicular to the track, got that part?**

Then the wheels of the car would match the grooves and the friction would be close to zero because they are coated with super teflon.

The front and back of the car would have bea ...[text shortened]... f acceleration?

We are not talking relativistic effects here, strictly say, below half of c. - 18 Jan '10 18:59Yep, something seems amiss here.

The question suggests that: (1) the power input; (2) the acceleration; and (3) the mass are all constant. This can't be the case unless time stands still.

Using the kinetic energy approach, we start with:

KE = (1/2)mv^2

The change in kinetic energy over time is equal to the power input, which in this case is a constant equal to 1 MW. So we have:

dKE/dt = P = d((1/2)mv^2)/dt)

Integrating both sides, and assuming the engine starts from rest such that (1/2)mv(0)^2 = 0, we have:

int(P,dt) = int(1, d(1/2)mv^2)

Pt = (1/2)mv^2

Solving for m, we have:

m = 2Pt/v^2

To make things more clear, we'll convert the velocity into an equivalent expression using acceleration:

a = dv/dt

int(a,dt) = int(1,dv)

Assuming both v(0) = 0 and a(0) = 0, we have:

at = v

Subbing this into the expression for "m" we obtained above, we have:

m = 2P/ta^2

Now, if: (1) the power input "P"; (2) the acceleration "a"; and (3) the mass "m" are all constant, we are still left with the variable "t". This means that all these quantities can only be constant if "t" is constant as well, indicating a snapshot in time. If we assume both: (1) the power input; and (2) the mass are constant, while the acceleration changes, we can solve for "v" using the first expression above:

v^2 = 2Pt/m

v = SQRT(2Pt/m) = SQRT(2P/m)*SQRT(t)

Taking the derivative to determine "a", we have:

dv/dt = a = SQRT(P/2m)/SQRT(t)

As with just about any inverse relationship, the limit of "a" as you approach t = 0 from the right side does not exist, as the function shoots off into infinity. That means that the proposed imaginary engine will give a ship of any particular mass an acceleration that passes through a = 9.81 m/s^2. This behaviour doesn't seem physical, but that is usually he case in idealized problems (for better or for worse). Therefore, as posed, I think the answer to the question must be "any size spacecraft."

(This seems like one of those questions that would benefit immensely from additional stipulations or clarification...) - 18 Jan '10 19:32

Maybe I'm missing something...*Originally posted by PBE6***Yep, something seems amiss here.**

The question suggests that: (1) the power input; (2) the acceleration; and (3) the mass are all constant. This can't be the case unless time stands still.

Using the kinetic energy approach, we start with:

KE = (1/2)mv^2

The change in kinetic energy over time is equal to the power input, which in this case is a const ...[text shortened]... would benefit immensely from additional stipulations or clarification...)

Without friction ANY mass can be moved by a constant accelleration.... - 18 Jan '10 19:52

True, in a frictionless scenario you just have to apply a constant force and you get a constant acceleration since F = ma. However in the case of a constant power input, the force and the velocity must act in a reciprocal manner since P = Fv. If we want to hold the force constant in order to get a constant acceleration, that means P will vary proportionally with v. However, this is contrary to the assumption in the question that the power input is constant. The only case where constant power input, constant acceleration and (necessarily) constant velocity coexist is when P = 0 and F = 0, again contrary to the assumptions in the question.*Originally posted by TheMaster37***Maybe I'm missing something...**

Without friction ANY mass can be moved by a constant accelleration....

So I agree with you, the answer should be "any mass". I think sonhouse was looking for a specific mass though, which makes me think the question has a bug in it somewhere. - 18 Jan '10 22:40 / 2 edits

The answer is NOT "any mass". There is NO solution to problem as posted.*Originally posted by PBE6***True, in a frictionless scenario you just have to apply a constant force and you get a constant acceleration since F = ma. However in the case of a constant power input, the force and the velocity must act in a reciprocal manner since P = Fv. If we want to hold the force constant in order to get a constant acceleration, that means P will vary proportionally w oking for a specific mass though, which makes me think the question has a bug in it somewhere.**

by definition v=at

substitute for v in Energy equation (m/2)v^2 = Pt

(m/2)(a^2)t = P

we are told a and P are constant so the mass must be inversely proportional to t.

ie it starts with infinite mass and dcreases to zero!!!

We deduce that constant power output does not give constant acceleration. - 19 Jan '10 01:32

So how much energy does it take to make say, 1 kg, accelerate at one G?*Originally posted by wolfgang59***The answer is NOT "any mass". There is NO solution to problem as posted.**

by definition v=at

substitute for v in Energy equation (m/2)v^2 = Pt

(m/2)(a^2)t = P

we are told a and P are constant so the mass must be inversely proportional to t.

ie it starts with infinite mass and dcreases to zero!!!

We deduce that constant power output does not give constant acceleration.

It's not rocket science for god's sake

No, wait....