Posers and Puzzles
18 Jan 10
19 Jan 10
Originally posted by AThousandYoungOr, if you prefer:
Pt = K
(m d^2/t^3)(t) = (m d^2/2t^2)
(m d^2/t^3)(t) = 1/2 m(at)^2
2(m d^2/t^3)(t)/(at)^2 = m
2P/ta^2 = m
See? The mass depends on the time given constant power and acceleration.
Pt = K
Pt = mv^2/2
P/t = mv^2/2t^2
P/t = ma^2/2
2P/ta^2 = m
- Joined
- 10 Dec 06
- Moves
- 8528
19 Jan 10
Originally posted by AThousandYoung"one megawatt applied for one second can give you how much mass the thing can accelerate for one second(sonhouse)"
Or, if you prefer:
Pt = K
Pt = mv^2/2
P/t = mv^2/2t^2
P/t = ma^2/2
2P/ta^2 = m
This is the problem he most recently asked, he gives a time of one second, an acceleration of 1 G and a Power of 1 Megawatt.
19 Jan 10
Here's an interesting question.
Since P, a and m cannot all be constant, if P and m are constant the rate of acceleration will slow down proportionate to the square root of the velocity.
Is there an asymptote? Is there a point at which there is no acceleration without more power? What value is it?
- Joined
- 10 Dec 06
- Moves
- 8528
19 Jan 10
1 edit
Originally posted by AThousandYoungyes yours is the same as mine.
Or, if you prefer:
Pt = K
Pt = mv^2/2
P/t = mv^2/2t^2
P/t = ma^2/2
2P/ta^2 = m
And sorry, I cannot partake in this anymore. its mid-night here and I have a formal lab paper due tomorrow...which i will now have to stay up all night to do.
😞
19 Jan 10
Originally posted by AThousandYoungNo
Here's an interesting question.
Since P, a and m cannot all be constant, if P and m are constant the rate of acceleration will slow down proportionate to the square root of the velocity.
Is there an asymptote? Is there a point at which there is no acceleration without more power? What value is it?
If there is a power you are putting energy into the system.
in other words 1/2 m v^2 is increasing
therefore v is increasing
therefore there is acceleration no matter what the mass
or what the power
or what the velocity
(Ignoring relativistic effects as previously stated)
Perhaps an interesting problem is what is the equation of motion
(as a function of t) for a constant mass with a constant power supply?
19 Jan 10
5 edits
Originally posted by wolfgang59T=square root of 2S/A, that says how much time it takes to go a certain distance under a certain acceleration. You don't need mass in this version. This is just using the famous S=(A*(T)^2)/2 where S= distance in whatever units, T is time in seconds, A is acceleration in the same units as S and solving for T as in the first equation. Simple stuff really.
No
If there is a power you are putting energy into the system.
in other words 1/2 m v^2 is increasing
therefore v is increasing
therefore there is acceleration no matter what the mass
or what the power
or what the velocity
(Ignoring relativistic effects as previously stated)
Perhaps an interesting problem is what is the equation of motion
(as a function of t) for a constant mass with a constant power supply?
For planetary navigation purposes that doesn't do much good because using that formula you arrive at your destination doing maximum velocity. It seems on the larger order of things you would want to be doing zero relative velocity at the end of the journey, but hey, what do I know🙂
So the correct formula, where you agree beforehand to start off using constant acceleration but only to the halfway point and turn the thrusters around and decel the rest of the way, which of course takes somewhat more time but you don't end up passing by your planet of choice at 2000 miles per second or whatever🙂
So that reworking is this:
T=2*square root of S/A
Using miles and feet, going 100,000,000 miles, like a journey to mars, under the first equation, it takes 2.102 and change days, arriving doing about 6 million feet per second, (1100 miles per second) and the second version, it takes 2.97 days and your average velocity half the first trip, or about 550 miles per second. And you arrive at zero relative velocity at Mars so you can actually think about landing🙂
19 Jan 10
Originally posted by sonhouseYou are assuming constant acceleration and deceleration which as you point out is a trivial problem.
T=square root of 2S/A, that says how much time it takes to go a certain distance under a certain acceleration. You don't need mass in this version. This is just using the famous S=(A*(T)^2)/2 where S= distance in whatever units, T is time in seconds, A is acceleration in the same units as S and solving for T as in the first equation. Simple stuff really.
...[text shortened]... d. And you arrive at zero relative velocity at Mars so you can actually think about landing🙂
My question was
Perhaps an interesting problem is what is the equation of motion
(as a function of t) for a constant mass with a constant power supply?
We have from previous posts that v^2 = 2P/mt
so v ={2P/m}^0.5 * t^-0.5
define K={2p/m}^0.5
therefore v = Kt^-0.5
v of course is dx/dt so to X as a function of t we need to integrate the above
Any takers?
19 Jan 10
Originally posted by wolfgang59That is interesting. However, I note nobody has answered my question which I seem to have to restate as this: how much mass can an accelerating body have on it and achieve one standard G of accel being accelerated with the power of one megawatt continuously applied, where we are talking 100 % conversion of that power into acceleration. I don't think anyone got that right yet.
You are assuming constant acceleration and deceleration which as you point out is a trivial problem.
My question was
[b]Perhaps an interesting problem is what is the equation of motion
(as a function of t) for a constant mass with a constant power supply?
We have from previous posts that v^2 = 2P/mt
so v ={2P/m}^0.5 * t^-0.5
define K={2p/m ...[text shortened]... 5
v of course is dx/dt so to X as a function of t we need to integrate the above
Any takers?[/b]
19 Jan 10
Originally posted by sonhouseI'll give it to you again:
That is interesting. However, I note nobody has answered my question which I seem to have to restate as this: how much mass can an accelerating body have on it and achieve one standard G of accel being accelerated with the power of one megawatt continuously applied, where we are talking 100 % conversion of that power into acceleration. I don't think anyone got that right yet.
NO SOLUTION!
Because:
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 decreases to zero!!!
What part of that do you not understand or disagree with?
19 Jan 10
1 edit
Originally posted by sonhouseI gave you the answer. So did several others. We cannot give you a single numerical value because it varies with t somehow.
That is interesting. However, I note nobody has answered my question which I seem to have to restate as this: how much mass can an accelerating body have on it and achieve one standard G of accel being accelerated with the power of one megawatt continuously applied, where we are talking 100 % conversion of that power into acceleration. I don't think anyone got that right yet.
2P/ta^2 = m
19 Jan 10
1 edit
Originally posted by sonhouseWhy do you think a constant power supply on a constant mass will give a constant acceleration? Besides "it's obvious" or "it seems to make sense"?
That is interesting. However, I note nobody has answered my question which I seem to have to restate as this: how much mass can an accelerating body have on it and achieve one standard G of accel being accelerated with the power of one megawatt continuously applied, where we are talking 100 % conversion of that power into acceleration. I don't think anyone got that right yet.