Posers and Puzzles
18 Jan 10
19 Jan 10
Originally posted by wolfgang59Correct. I worded the problem badly.
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?
Is there an asymptotic maximum velocity for every particular power level? I think there is because velocity only increases with the square root of the added energy. Thus if we plot velocity vs time we'll get a y=x^2 sort of curve, which is asymptotic I think.
19 Jan 10
2 edits
Originally posted by AThousandYoungJeez, it's already being done with ion rockets on voyages around the solar system as we speak. Why do you think it would be otherwise? You kick a ball with X amount of force, it accelerates X amount, you keep doing that the accel stays exactly the same. What is the problem here? The final velocity is independent of the force applied, it is just the increase in velocity that is directly related to the power applied.
Why 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"?
If what you are saying is true, then if you start out with a mass going one km/sec and hit it with X amount of power and compare that to the exact same mass going 1000 Km/sec and hit it with the same power, how does the mass know what velocity it is going, outside of relativistic effects?
Your argument sounds like the 19th century argument against rockets, they can't work because there is nothing to push against, not realizing it pushes against its own inertia.
19 Jan 10
Originally posted by sonhouseCan you show a mathematical analysis of these ion engines showing constant power, acceleration and mass?
Jeez, it's already being done with ion rockets on voyages around the solar system as we speak. Why do you think it would be otherwise? You kick a ball with X amount of force, it accelerates X amount, you keep doing that the accel stays exactly the same. What is the problem here?
Why do I think so? I showed you the math already.
19 Jan 10
Originally posted by sonhouseThe second mass contains more kinetic energy, no?
Jeez, it's already being done with ion rockets on voyages around the solar system as we speak. Why do you think it would be otherwise? You kick a ball with X amount of force, it accelerates X amount, you keep doing that the accel stays exactly the same. What is the problem here? The final velocity is independent of the force applied, it is just the increase ...[text shortened]... same power, how does the mass know what velocity it is going, outside of relativistic effects?
19 Jan 10
3 edits
Originally posted by AThousandYoungYes but it doesn't know that, it is only kinetic energy in relation to some other reference.. Are we confusing non-relativistic V relativistic effects here?
The second mass contains more kinetic energy, no?
There is conservation of energy here, the kinetic energy cannot exceed the power input no matter what you do. More accurately, the kinetic energy cannot exceed the total power input over the time period of interest.
19 Jan 10
Originally posted by sonhouseThat has always bugged me.
Yes but it doesn't know that, it is only kinetic energy in relation to some other reference.. Are we confusing non-relativistic V relativistic effects here?
It seems like it should take the same energy to fire a bullet as for the bullet to stay still and everything else move. But it doesn't...does it?
19 Jan 10
Originally posted by AThousandYoungof course not, but if the bullet happens to pass through another barrel that magically closes and the exact same energy again supplied, and it is in space, the change in velocity will be exactly the same as the first change in velocity. What is so hard to understand about that?
That has always bugged me.
It seems like it should take the same energy to fire a bullet as for the bullet to stay still and everything else move. But it doesn't...does it?
19 Jan 10
Originally posted by sonhouseI think I get it.
of course not, but if the bullet happens to pass through another barrel that magically closes and the exact same energy again supplied, and it is in space, the change in velocity will be exactly the same as the first change in velocity. What is so hard to understand about that?
An object's movement is only relative when you're talking about inertial frames. This is not an inertial frame of reference we're discussing.
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19 Jan 10
1 edit
Originally posted by sonhouseSonhouse I answered the question earlier when you put a finite time on it...but now you gone and rewarded the problem that we worked so hard to restate,and it once again has no solution.
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.
Here is the one that has a solution.
"one megawatt applied for one second can give you how much mass the thing can accelerate for one second(sonhouse)"
To which you can use
2(P*t)/(a*t)^2 = m
a = 1G
t=1 s
and P = 1MegaWatt
to give you
m = 20,824 kg
19 Jan 10
Originally posted by joe shmoWhat is the basic difference between a certain mass accelerated for one second by a certain amount of power and that same amount of power on a per second basis applied continuously? If I accelerate with one megawatt for one second and get one G and as you say can carry 20000 odd Kg with me, what happens in second two that is different from second one? or second 3 or 4?
Sonhouse I answered the question earlier when you put a finite time on it...but now you gone and rewarded the problem that we worked so hard to restate,and it once again has no solution.
Here is the one that has a solution.
"one megawatt applied for one second can give you how much mass the thing can accelerate for one second(sonhouse)"
To which you ...[text shortened]... use
2(P*t)/(a*t)^2 = m
a = 1G
t=1 s
and P = 1MegaWatt
to give you
m = 20,824 kg
19 Jan 10
Originally posted by AThousandYoungThe units being P=power, like watt/seconds? t=time, in seconds? a=acceleration in some unit/time^2 and m=mass as in kilograms?
Let's be rigorous here. Sonhouse, do you agree with my derivation that
P/ta^2 = m
is correct?
So the formula is m=(P/ta)^2? Not sure the exact formula
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19 Jan 10
1 edit
Originally posted by sonhouseBecause power is the rate of change of energy with respect to time, so in order for power to be useful in terms of energy it has to be applied for a definite length of time.
What is the basic difference between a certain mass accelerated for one second by a certain amount of power and that same amount of power on a per second basis applied continuously? If I accelerate with one megawatt for one second and get one G and as you say can carry 20000 odd Kg with me, what happens in second two that is different from second one? or second 3 or 4?
I mean a Power is applied for a certain length of time, and during that time the Power is transformed into energy, beit kinetic, potential,ect...It is onlt this energy that will cause an acceleration if we are talking about kinematics. Power do not do anything until it is converted to energy. Are we getting anywhere?
19 Jan 10
2 edits
Originally posted by sonhousePower would be watts, not watt/seconds, but yes.
The units being P=power, like watt/seconds? t=time, in seconds? a=acceleration in some unit/time^2 and m=mass as in kilograms?
So the formula is m=(P/ta)^2? Not sure the exact formula
No, the t and P are not squared.