From only the video footage within the capsule, please estimate the rocket's acceleration (gee's).


Show all work and assumptions.

There is a different but related video below, also with swinging objects, in case the original BBC link fails at some point in the future.

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    $\begingroup$ I have an estimate, but don't want to spoil it for others. $\endgroup$ Commented Apr 20, 2017 at 19:10
  • $\begingroup$ I suppose it's related to the period of swing of the stuffed doggy? Might be error introduced due to vibration. $\endgroup$
    – Steve
    Commented Apr 21, 2017 at 13:23
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    $\begingroup$ Here is a video not requiring obscure technologies like Flash: youtube.com/watch?v=3iPuB-Ruzi4 $\endgroup$ Commented Apr 22, 2017 at 17:55
  • $\begingroup$ @Hohmannfan I thought I have Flash blocked in my Chrome browser on OSX. Occasionally I'll have a BBC website blocked, but I thought it delivers video in an alternate way to me when I block flash. I'll have to look into it, thanks for pointing it out. The video is definitely a superior video and data source, thanks! $\endgroup$
    – uhoh
    Commented Apr 23, 2017 at 0:51

1 Answer 1


This video from the launch includes some footage from before liftoff were the pendulum has some momentum. I counted 8 periods over 10.2 seconds.

From that, we can find the length of the pendulum:

$$L = g\left(\frac{T}{2\pi}\right)^2$$


$$L \approx 40cm$$

51 seconds after launch, I can start to count again, giving 20 periods over 17.4 seconds.

$$g = \frac{L}{\left(\frac{T}{2\pi}\right)^2}$$


$$g \approx 21 m/s²$$

We have to subtract gravity, so the rocket is accelerating at a little more than $1.1g$.

At 1 minute 48 seconds (right before the escape tower goes), I can count 8 periods over 6.0 seconds. Using the same logic, that is about $1.85g$

Edit: I found another way to get a value for the acceleration: The camera angle in the beginning is constant, so we can use the time the rocket uses to accelerate its own height (45.6m). That is 4.9 seconds, so the initial acceleration is $0.39g$

At 2:57 a velocity indicator is visible, but that is cheating.

As experienced by the cosmonauts:

 -1s: 1.0g
 +2s: 1.4g
 51s: 2.1g
108s: 2.85g
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    $\begingroup$ @uhoh Thanks, I forgot that I was supposed to show what the cosmonauts experienced. Editing in the data now. $\endgroup$ Commented Apr 23, 2017 at 8:11
  • $\begingroup$ 51 seconds after launch the Soyuz will have done a significant part of its pitch program so you can't just subtract 9.8 m/s^2 from the acceleration experienced in the rocket's frame of reference to get the Earth-frame acceleration. You'd have to do a vector subtraction of Earth g from the acceleration vector measured in the Soyuz's frame. But I didn't see anything in the video that would give the Soyuz's orientation with respect to the local vertical, so though you can estimate the magnitude of that acceleration vector, you don't know its direction. $\endgroup$ Commented Jan 12, 2019 at 19:36

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