Posers and Puzzles
19 Jan 10
21 Jan 10
Originally posted by PBE6What he said.
The "actual acceleration" is a function of the velocity.
dKE/dt = P
Pt = KE = 1/2mv^2
Taking the derivative of both sides with respect to t:
P = mv*dv/dt
dv/dt = a = P/(mv)
Subbing in the values provided, we have:
a = (1)/((1)*v) = 1/v
The limit of "a" as v->0 from the right side does not exist as the function shoots off to infinity. Th ...[text shortened]... The acceleration, being a smooth curve, will take on all values in between at some point.
27 Jan 10
3 edits
Originally posted by joe shmoThe thing I am having trouble wrapping my mind around is when under acceleration by a constant power source, what does the mass know about the velocity? When you are under relativistic velocity, how does the mass know it's doing 1 km/sec V 1000 Km/sec?
how many times must we go over this???? YOU NEED TO STIPULATE A TIME......:'(
I don't know what your calculating, but its not the acceleration of the body in question given the parameters.
have a look at this
in this problem
acceleration = sqrt(2*power/(mass*time))
you have stipulated
Power = 1watt
mass = 1kg
time is required to find acc ...[text shortened]... elf that it is indeed a curve, and NOT A HORIZONTAL LINE(meaning acceleration is NOT CONSTANT)
It sounds like there is an impedance to further velocity change that is velocity driven and I can't see how that happens.
Well here is a link that seems to support what I am saying but there is only the one statement and no math except as a lead-in to relativistic acceleration:
Acceleration by Constant Force
Posted September 23, 2006 7:53 AM by Jorrie
Pathfinder Tags: acceleration relativity speed of light
In Newton's mechanics, if we apply a constant force to an object of constant mass, the acceleration will be constant. In principle we can accelerate such an object to exceed the speed of light. Not so in relativistic mechanics. The following snippet out of the eBook Relativity 4 Engineers (the link refers to a special offer to CR4 readers) shows how to compute the acceleration for a relativistic scenario: (note that the symbol v here means 'normalized velocity', i.e., v=v/c, where v is velocity in m/s and c is the speed of light in m/s).
Here is another link with math that supports the same conclusion, constant force=constant acceleration:
http://www.euclideanspace.com/physics/kinematics/acceleration/index.htm
27 Jan 10
Originally posted by sonhouseNobody here will disagree that constant force equals constant acceleration.
The thing I am having trouble wrapping my mind around is when under acceleration by a constant power source, what does the mass know about the velocity? When you are under relativistic velocity, how does the mass know it's doing 1 km/sec V 1000 Km/sec?
It sounds like there is an impedance to further velocity change that is velocity driven and I can't see ...[text shortened]... nt acceleration:
http://www.euclideanspace.com/physics/kinematics/acceleration/index.htm
however the original (and subsequent) post talks about constant power
Force, power, energy are all different.
I must say that your last post made me think though. What the equations show is that given a constant power source aboard our imaginery space-ship we can calculate our velocity by measuring the acceleration!
27 Jan 10
Just for fun (insert nerd joke here), I decided to see what the power function looks like for constant acceleration/constant force.
From previous results, we have:
P = Fv
Taking the derivate of both sides with respect to time, and keeping the force constant, we have:
dP/dt = F*dv/dt = Fa
From Newtown's second law, it follows that:
F = ma
a = F/m
Subbing this into the expression above, we have:
dP/dt = F^2/m
Integrating this expression with respect to time, we get:
P = (F^2/m)*t
Therefore, the amount of power supplied by the engine must increase linearly with time to maintain a constant acceleration.
27 Jan 10
4 edits
Originally posted by wolfgang59Given a constant power source, or a variable source, you can still calculate your resultant velocity if you know the acceleration history. I am still having trouble wrapping my head around the need for more power to accelerate the same mass that is say, now at 1000Km/second versus the same mass at 1 Km/sec. What is the physical basis for that? Why does the force go down with velocity? If you are situated say, between stars 10 light years away from any other mass, how does the system know that the velocity is now 1000Km/sec not 1?
Nobody here will disagree that constant [b]force equals constant acceleration.
however the original (and subsequent) post talks about constant power
Force, power, energy are all different.
I must say that your last post made me think though. What the equations show is that given a constant power source aboard our imaginery space-ship we can calculate our velocity by measuring the acceleration![/b]
The thing that continues to bug me is this is way below any relativistic effect velocity.
I assume all this means that if I have 1 hp, 746 watts which is defined as the power required to lift 555 pounds against 1 gravity one foot in the air, which really means an acceleration of 1/32 of a G, that starting at zero velocity somewhere in space, you start out some acceleration, say 32 hp against 555 pounds which to me says that should result in an acceleration of 1 full G, that somehow the force goes down simply because at some later time the velocity is higher? So at some point the actual acceleration keeps reducing till it is for all intents and purposes zero? I assume if that is true it would be asymptotically approaching zero but never actually making it. What I am having a devil of a time with is understanding how the system knows it is at a different velocity from one time frame to the next.
I made a statement answering ATY when he said you get X amount of velocity kicking a bullet of a given mass with X amount of explosive force where the acceleration ceases very shortly after leaving the barrel and I said I thought if a second barrel was to show up magically and closed off very quickly after the bullet entered and say the first gun gave it a velocity of 1000 feet per second or whatever units you choose, and then the bullet goes to a second 'gun' which closes off exactly at the right time and has another mass of explosive of exactly the same power that now this time the bullet does not in fact exit barrel # 2 at 2000 feet per second? But now because of this to me mysterious effect, it only exits barrel # 2 at something less, say 1800 feet per second, NOT based on some friction effect but on the mass itself?
I found this piece by Professor Taylor of Oberlin college:
https://kb.osu.edu/dspace/bitstream/1811/2458/1/V30N04_218.pdf
He gives all the equations of motion from constant force and constant power.
I just have a hard time visualizing why that is, I still don't see the connection between how the mass knows it is at say 1 Km/second at the start and 1000 Km/sec at the end of an acceleration run. (in space far from any other mass)
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27 Jan 10
Originally posted by PBE6This I like...Its mathematically and phyically satisfying that constant power does not equal constant acceleration
Just for fun (insert nerd joke here), I decided to see what the power function looks like for constant acceleration/constant force.
From previous results, we have:
P = Fv
Taking the derivate of both sides with respect to time, and keeping the force constant, we have:
dP/dt = F*dv/dt = Fa
From Newtown's second law, it follows that:
F = ma
a ...[text shortened]... er supplied by the engine must increase linearly with time to maintain a constant acceleration.
27 Jan 10
3 edits
Originally posted by joe shmoSo I think I better redefine my basic assumptions, like I had at one point worked out that since one horsepower is defined as = 746 watts, that would have to be continuously applied to be defined as a power function. Show me my errors if I am wrong. Further, I took one horsepower also defined as the power to move 555 pounds one foot in one second. So that is the same as one horsepower moving 555 pounds STRAIGHT UP one foot in one second. That is the same as saying one horsepower can accelerate 555 pounds at 1/32 of a G, so 32 times that would be saying 32 horsepower can accelerate 555 pounds at one G. I took that to mean that is what you would get if you were in a spacecraft that weighed 555 pounds on earth and it was placed into space somewhere away from the influence of other masses and if you applied 32 HP to 555 pounds you would get a constant 1 G of acceleration. Is that now not correct?
This I like...Its mathematically and phyically satisfying that constant power does not equal constant acceleration
I guess the answer to that is no.
Given that, what about a constant thrust? Isn't that the same thing? If you say you have a 1000Kg spacecraft under a thrust of 1000Kg, you get 1 G of acceleration, right? Are we now also saying that same thrust say ten seconds later will not be giving the craft 1 G of acceleration but some lesser amount? This would be assuming there is no significant change of total mass.Is that the same thing as saying a certain amount of power applied?
For instance, I am thinking in that case of a laser sail, where a massive laser beam is concentrated on a large reflector attached to a spacecraft and it gets a constant thrust more or less, as it leaves the earth, in this case if we assume the laser power goes up just to make up for inverse square losses of laser power with distance, that is not enough to keep the acceleration of the craft constant?
27 Jan 10
Another nerd moment - just to compare, I thought I'd examine constant power input.
P = dKE/dt
Pdt = dKE
If power is constant, then we have:
Pt = KE = (1/2)mv^2
And the kinetic energy increases linearly. Solving for v, we get:
v = SQRT(2Pt/m) = SQRT(2P/m)*SQRT(t)
This result tells us that the velocity is only proportional to the square root of t, and thus increases much more slowly than linear velocity achieved with a constant acceleration. Further, taking the derivate of this expression we have:
dv/dt = a = SQRT(2P/m) / (2*SQRT(t)) = SQRT(P/2m) / SQRT(t)
This acceleration curve starts with a discontinuity at t=0 which is an artifact of the assumption that P=k at t=0, but otherwise shows that the acceleration slows significantly as the spaceship achieves higher velocities.
It's interesting (again, insert nerd joke here) to compare the two situations (A) constant acceleration; and (B) constant power input on a variable-by-variable basis:
Power Input
(A) increases linearly with time
(B) remains constant
Kinetic Energy
(A) increases quadratically with time
(B) increases linearly with time
Acceleration
(A) remains constant
(B) decreases with time
Velocity
(A) increases linearly with time
(B) increases proportionally to the square root of time
Position
(A) increases quadratically with time
(B) increases proportionally to the cube of the square root of time (3/2 power)
As you can see, the constant acceleration method (A) provides a quicker trip, but at the cost of increasing power consumption. The kinetic energy and velocity comparisons are especially telling, as they illustrate how a linear increase in velocity requires a quadratic increase in kinetic energy. This notion is sometimes counter-intuitive, especially when it is confused with momentum as in the example link sonhouse provided. Fascinating!
OK, enough nerding out for one night.
28 Jan 10
Originally posted by PBE6So with all that in mind, lets get back to the article about the VASIMIR ion rocket. You remember the piece where they said with a quote 200 megawatt unquote power supply, you can go from Earth to Mars in 39 days? I and ATY I think it was, calculated the acceleration as being around 4 milliG's but that was a constant acceleration, an average. So what is the beginning acceleration and ending acceleration, or is that not possible to compute not knowing the mass of the spacecraft?
Another nerd moment - just to compare, I thought I'd examine constant power input.
P = dKE/dt
Pdt = dKE
If power is constant, then we have:
Pt = KE = (1/2)mv^2
And the kinetic energy increases linearly. Solving for v, we get:
v = SQRT(2Pt/m) = SQRT(2P/m)*SQRT(t)
This result tells us that the velocity is only proportional to the square root ...[text shortened]... in the example link sonhouse provided. Fascinating!
OK, enough nerding out for one night.
28 Jan 10
1 edit
Originally posted by sonhouseIt wasn't me. I'm thinking about it now. I think it needs to be broken up.
So with all that in mind, lets get back to the article about the VASIMIR ion rocket. You remember the piece where they said with a quote 200 megawatt unquote power supply, you can go from Earth to Mars in 39 days? I and ATY I think it was, calculated the acceleration as being around 4 milliG's but that was a constant acceleration, an average. So what is the ...[text shortened]... ending acceleration, or is that not possible to compute not knowing the mass of the spacecraft?
First of all the energy required to get out of Earth's gravity plus power tell you how long that takes.
Then you have a ship at zero velocity just outside of Earth (can I make this simplification or do I need to figure out orbital velocity and crap?)
Then you accelerate for the first half and decelerate for the second half. This should be a symmetric pattern so we can analyze constant power causing acceleration for half the distance. Then double those numbers for the deceleration.
There will not be a constant acceleration; the time can probably be determined without a constant acceleration.
28 Jan 10
1 edit
Originally posted by AThousandYoungOops, the Vasimir does not go from Earth to orbit. Forget that first part.
It wasn't me. I'm thinking about it now. I think it needs to be broken up.
First of all the energy required to get out of Earth's gravity plus power tell you how long that takes.
Then you have a ship at zero velocity just outside of Earth (can I make this simplification or do I need to figure out orbital velocity and crap?)
Then you acceler ...[text shortened]... a constant acceleration; the time can probably be determined without a constant acceleration.
So really it's just a question of "given constant power and mass, how long will it take to go halfway?"
28 Jan 10
2 edits
Originally posted by AThousandYoungI already worked out what the G force would be but that was based on my assumption of constant acceleration for a constant thrust. I got about 4 milliG's, if you do a constant 4 milli G's from Earth orbit, to halfway to mars, say the total trip is 100 million miles or 160 million kilometers, you go halfway, turn the thrusters around and decel the rest of the way leaving you in the vicinity of Mars with something like zero relative velocity. That was predicated on the power mentioned in the article of 200 megawatts, which would have to have been done with some sort of nuclear reactor. Anyway, it specified a constant power source of 200 megawatts. But according to these new laws, (it appears to have been realized in 1930 by Lloyd W Taylor at Oberlin College), at least he claims in the paper not to have seen constant power equations before, but with a constant accel of 0.04 G's you get to around 300 miles per second max velocity at midway, something like that but if the guy who made the pronouncement of a 39 day trip did not take variable G forces into account, maybe his timeline is wrong. I guess we could just make up an assumed mass and go from there, say the whole thing has a mass of 10,000 kg and go from there with the constraint of a 39 day trip or 19.5 day (1.6848E6 seconds) to the halfway point, what would the min and max G force be?
Oops, the Vasimir does not go from Earth to orbit. Forget that first part.
So really it's just a question of "given constant power and mass, how long will it take to go halfway?"
Maybe you can solve for mass given a 19.5 day halfway trip with 200 megawatts and see if the mass looks reasonable. For instance, if the mass ends up being 200 Kg, we can be reasonably sure there is something wrong in Denmark, since there is no way to make a 200 meg PS that weighs only a couple hundred pounds.
The Taylor paper has some equations, like A= Sqr root (P/2MT) and
S (distance) = sqr root ((8P*(t^3)/9M)) Can anyone verify these equations? They are on the PDF file of the Taylor paper, but I don't know what units exactly to use. I think in metric talk, power is in ergs, but what about mass, grams or Kilograms? Distance in meters? Cm? Time for sure in seconds. My first attempt to use 10,000 Kg directly and ergs gave me a distance of about 8 million units (kilometers? Meters? Centimeters?, not sure) at any rate the distance did not work out to anything like a trip to Mars so I clearly did something wrong. My assumption about power put 200 megawatts times 1E7 to make it ergs. I think that's right, ten million ergs=1 watt/second. Still no joy.