Alright, so i've been thinking about putting together a calc on the Tria decelartion incident in "The Return Part 1". Problem is, there are a number of unknown variables in that incident which make it quite hard to do so, in any case, i thought you might provide some help with that. To be perfectly honest, i'm only doing this to get a ballpark figure of what we saw, i'm not expecting anything even remotely accurate.
First off, i believe we need to look no further then the standard KE=(1/2)*m*v^2 formula to calculate this. So the unknown then is mass, because velocity is said to be 0.999c, which also happens to be 299490210 m/s.
So, mass, any suggestions on what that might be? I haven't even got a clue on how big the Aurora-Class is supposed to be, except that it seems larger then the BC-304 in every shot we have the two in close Proximity. So for this preliminary calc, i'm going with 1km. Preliminary meaning that i'm going to change it and punch in the correct numbers when i get your input on it.
So, a 1km ship, that still doesn't give us a mass. I'm unsure of how to proceed from here, so again, input will be welcome (and required, damnit). But i've seen calcs like this done before, where the mass of the Nimitz-Class Carrier has been used as a standard, all you have to do is gauge the size difference, so for now at least, i'm proceeding like that. So, a Nimitz is quite flat and a little over 300m long, meaning, what, the Aurora is about 12 times larger?
Ok, on to the preliminary calc (that will be changed with input). KE = (1/2)*(12x102000000kg)*(299490210 m/s)^2. This gives us the value of 5.489e25J, in other words, that for the whole breaking manouver.
But to get this into Watts, we need to factor in the time, now this is another tricky bit, because we get a figure of 27g. However, if the Tria would be breaking at 27g the entire time, then it would take almost two weeks for it to come to a complete stop. However, the goal was not to get the Tria to stop completly, only match speeds with the BC-304 that was going as fast as it could. But in whatever way we look at this, we are getting extrodinary long waiting times, which definetly doesn't match Caldwell's "a few hours untill we match speeds" estimate. So i'm guessing the 27g figure was just the initial breaking. When the Tria found out that the Deadalus was willing to escort them, they applied the real breaks, so to speak.
But this still leaves us without a figure. If it would take about a few hours to match speeds with the Deadalus, then it would be travelling at extreme speeds (relativistic speeds) if the 27g figure is right, but i hardly see that as possible. In any case, to be generous, let's assume the Deadalus was travelling at half that speed (which is an insane speed), meaning that if it took "a few hours" to match speeds it would have taken "a few hours" more to come to a complete stop. Ah heck, for this preliminary calc, let's just call it a day shall we. I mean it's still going to be changed around a lot so nothing's final yet, i've just brought up the points that need to be adressed.
One day means 86400 seconds. This in turn gives us a Wattage of 6.35e20W.
Ok then, now that i've completed my utterly faulty and preliminary calc i'm going to ask people to chip in so we can come to a figure closer to the truth. In any case, this will never be an accurate calc, at best it will give us a ballpark figure that can be several orders of a magnitude wrong. But. I'd still like to try.
So, get cracking people, what i want is the real time involved and the real mass involved, or figures as close as possible to those.
First off, i believe we need to look no further then the standard KE=(1/2)*m*v^2 formula to calculate this. So the unknown then is mass, because velocity is said to be 0.999c, which also happens to be 299490210 m/s.
So, mass, any suggestions on what that might be? I haven't even got a clue on how big the Aurora-Class is supposed to be, except that it seems larger then the BC-304 in every shot we have the two in close Proximity. So for this preliminary calc, i'm going with 1km. Preliminary meaning that i'm going to change it and punch in the correct numbers when i get your input on it.
So, a 1km ship, that still doesn't give us a mass. I'm unsure of how to proceed from here, so again, input will be welcome (and required, damnit). But i've seen calcs like this done before, where the mass of the Nimitz-Class Carrier has been used as a standard, all you have to do is gauge the size difference, so for now at least, i'm proceeding like that. So, a Nimitz is quite flat and a little over 300m long, meaning, what, the Aurora is about 12 times larger?
Ok, on to the preliminary calc (that will be changed with input). KE = (1/2)*(12x102000000kg)*(299490210 m/s)^2. This gives us the value of 5.489e25J, in other words, that for the whole breaking manouver.
But to get this into Watts, we need to factor in the time, now this is another tricky bit, because we get a figure of 27g. However, if the Tria would be breaking at 27g the entire time, then it would take almost two weeks for it to come to a complete stop. However, the goal was not to get the Tria to stop completly, only match speeds with the BC-304 that was going as fast as it could. But in whatever way we look at this, we are getting extrodinary long waiting times, which definetly doesn't match Caldwell's "a few hours untill we match speeds" estimate. So i'm guessing the 27g figure was just the initial breaking. When the Tria found out that the Deadalus was willing to escort them, they applied the real breaks, so to speak.
But this still leaves us without a figure. If it would take about a few hours to match speeds with the Deadalus, then it would be travelling at extreme speeds (relativistic speeds) if the 27g figure is right, but i hardly see that as possible. In any case, to be generous, let's assume the Deadalus was travelling at half that speed (which is an insane speed), meaning that if it took "a few hours" to match speeds it would have taken "a few hours" more to come to a complete stop. Ah heck, for this preliminary calc, let's just call it a day shall we. I mean it's still going to be changed around a lot so nothing's final yet, i've just brought up the points that need to be adressed.
One day means 86400 seconds. This in turn gives us a Wattage of 6.35e20W.
Ok then, now that i've completed my utterly faulty and preliminary calc i'm going to ask people to chip in so we can come to a figure closer to the truth. In any case, this will never be an accurate calc, at best it will give us a ballpark figure that can be several orders of a magnitude wrong. But. I'd still like to try.
So, get cracking people, what i want is the real time involved and the real mass involved, or figures as close as possible to those.
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