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Apollo 11 liftoff acceleration function 
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#1
Dec1713, 04:31 PM

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Hello,
Recently I saw a youtube video of the Apollo 11 launch with time and velocity data. I took data in 5 second increments and put the data in my graphing calculator. I used a quadratic regression to come up with a velocity time function. I then differentiated this function to come up with an acceleration time function, and for the first 22 seconds I have a negative acceleration. Where did I go wrong? My velocity function from quadratic regression: v(t) = 0.1368t^2  5.9846t + 102.863 Then differentiating for a(t): a(t) = 0.2736t  5.9846 I tried removing the point (0,0) so I could do an exponential regression, but this fits the data worse and gives a nonzero acceleration at t=0... So either I am using the wrong regression type or the data was bad. Is the launch simply too complex to fit into a nice differentiable function like this? I just thought it would be cool to have a handle on the types of acceleration experienced during this historic achievement. If you'd like me to post the data I would be happy to. Thanks in advance, Lee 


#2
Dec1713, 04:38 PM

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#3
Dec1713, 04:45 PM

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#4
Dec1713, 04:59 PM

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Apollo 11 liftoff acceleration function



#5
Dec1713, 05:18 PM

P: 63

Lee 


#6
Dec1713, 05:42 PM

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I recall something like 7 million pounds of thrust on a 6 million pound Saturn V rocket full of fuel. The initial launch acceleration would be fairly slow, 1/6 g upwards.



#7
Dec1713, 08:08 PM

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You give formulas for v(t) and a(t), but the independent variable is 'x'. Shouldn't 'x' be 't'?



#8
Dec1713, 09:16 PM

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I wonder if the actual function itself was changing during the launch, making the concept of a single function invalid. My data went from t=0 until t=160 seconds because at that point the first stage rocket is jettisoned and there is an abrupt change in the velocity. I wonder if shortening the timespan would improve the function. I know usually the more data the better, but if it does, it might indicate that the data should not be considered as one function, but several. Lee 


#9
Dec1713, 11:44 PM

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During stage separations, the rockets in the jettisoned stage have been cut off and the rockets in the next stage have not ignited, so the rocket is not accelerating during these intervals, at least due to thrust. There is a change in mass though from the loss of the stage being jettisoned.
By not including any data points during the first stage burn, you have necessarily lost accuracy in crafting your velocity and acceleration functions. 


#10
Dec1813, 12:36 AM

P: 63

Lee 


#11
Dec1813, 12:46 AM

P: 26

From what I understand, the equation that was created cannot be the right equation simply because of the y intercept first of all being more than 0.
So either the data was entered incorrectly or the regression function is not working properly. Perhaps run the regression in an excel spreadsheet vs a graphing calculator and try again. I would think that a rocket is just a simple object accelerating upwards, like those ball toss problems that physics 101 text books use. Here the ball is the rocket. However there might be something to consider, since the rocket is consuming fuel very quickly the change in mass is taking place, its getting lighter and the accelerating forces on a changing mass might cause the function to approx incorrectly, as the force acted upon the rocket remains the same. In other words I think since the rocket is producing I would assume consistent thrust per time unit, but accelerating a changing mass, that might be a problem. 


#12
Dec1813, 01:02 AM

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Lee 


#13
Dec1813, 02:30 AM

P: 26

But then I realized that the rockets acceleration is not constant unlike the ball, which is 9.8/ms^2, so yeah they are different, but I think that the derivative of acceleration of the right eq, would give us how fast the acceleration changes, but still not sure how to help solve the problem. 


#14
Dec1813, 02:36 AM

P: 26

perhaps, Im not sure but maybe a group of functions needs to be created, each unique function for maybe i dont 1 sec time intervals and plot each function on a graph and then approx one function from that, not sure if that makes sense, just something i came up with



#15
Dec1813, 03:20 AM

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Edit: And yet, that seems to be what they have gone with. I'm getting TWR of 1.23 from data on Wiki. I'll have to take a look at how that changes as fuel is burnt off, but this just doesn't seem like smart engineering, which really shouldn't be the case. Unless, increase in payload requirements came in late, and they had to increase the size of first stage tanks to compensate. But even then, some boosters would do a much better job of it... Weird. Edit 2: Curiouser and curiouser. I just plugged in some numbers. At the listed thrust of 34MN and I_{SP} of 263s, in the 150s that the first stage was firing, it'd burn through just under 2,000T of fuel. Which is consistent with masses listed. Plugging that into the rocket formula, I get total [itex]\Delta V[/itex] of 3,160m/s. But hanging out in Earth gravity for 150 seconds brings it down to 1,690m/s. Now, suppose I just poured out 600T of fuel. I really want that higher TWR on liftoff. No modifications to the rocket. We just reduce initial mass of the rocket, keeping final mass, after stage burns through, unchanged. Naturally, we get much less [itex]\Delta V[/itex]. Just 2,540m/s, in fact. But we also burned through all this fuel in just under 105 seconds instead of 150. After we take gravity's toll out, we're down to 1,515m/s. I don't know about you, but 175m/s for almost a third of your fuel seems a bit wasteful. And that's without taking into account any savings you can make by reducing tank size. Now, without taking drag into effect, this does buy you about 20km of extra altitude. That's significant on the ascent stage. Difference between 60km and 80km when you fire second stage would be huge. Of course, rocket would be far into rotation by then. But I have a feeling that once I add even a simple atmospheric model, I'm going to lose even that advantage. I'm going to go check that now. Edit 3: Alright, so even with atmosphere in place, it looks like this extra fuel is buying altitude. None of the above takes into account the fact that I_{SP} of the engines improves dramatically up there, but the idea is the same either way. Get as much fuel to thin atmo as you can, and let it all burn there. The whole thing could still be more efficient with a bit more thrust at the liftoff, but I'm guessing it comes down to working with what you have. I'm glad I got that sorted. Maybe it will be interesting to someone else. 


#16
Dec1813, 08:06 AM

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Rockets are hugely inefficient in the initial stages (zero efficiency for the first second or so). You just have to put up with virtually chucking away most of your fuel during and just after liftoff. Space engineers are constantly looking for better ways of launching but nothing viable has yet turned up. I am really surprised that aircraft launching is not yet the thing to use.



#17
Dec1813, 08:57 AM

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http://en.wikipedia.org/wiki/Delta_IV http://en.wikipedia.org/wiki/Antonov_An225_Mriya From the articles, you can see that the Delta IV launch system weight can easily exceed the maximum payload capacity of the An225 aircraft. Now the Russians have lately become interested in using the An225 in an air launch project: http://en.wikipedia.org/wiki/Air_launch_to_orbit Who knows what will happen? 


#18
Dec1813, 10:48 AM

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Also, another thing that you might not be considering here is the relative cost  rocket engines are very expensive, but rocket fuel is (relative to the cost of a launch) cheap. The fuel cost is a tiny percentage of the overall cost of a modern launch vehicle, so if you can gain any benefit by increasing fuel load without increasing engine count or thrust, even if it is hugely wasteful, it is probably worthwhile. This is especially true with liquid fuel rockets  if you look at the liftoff TWR of rockets powered only by liquid fuel, they tend to be pretty slow off the pad, while ones with solid rocket boosters tend to jump off the pad much faster (though still not typically above 1.5 or 1.6:1 or so) 


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