2016-03-12

# Speed Up Factor

I watch a lot of Coursera videos and usually view them at 1.25x or 1.5x normal viewing speed. I started thinking about how much that would translate into viewing time.

Factor  Time(s)
1.25x     75
1.00x    100
0.75x    125

From the table, my assumptions can be seen. Viewing a video at 1x speed should show the video in real-time. If the video is 100 seconds long, I would expect it to take 100 seconds at a speed up factor of 1x. If however, I choose to view the video at 1.25x speed up factor I would expect the video to be shorter, around 75 seconds. If the speed up factor were set to 0.75x, I would expect the video to take longer to watch, in the neighborhood of 125 seconds.

If $$t$$ is the time in seconds of the video at normal speed. We can formulate the time when a speed up factor is applied. Set $$t = 100$$, to simplify calculations and assumptions.

$\large 100\times 0.75 = 75$

This isn’t what we wanted.

$\large \frac{100}{0.75}\approx 133$

This isn’t what we wanted either! If we think of the speed up factor in terms of percentages we could try something like this:

$\large 100\times (1 - 0.75) = 25$

We are getting closer, this tells us how much time will be gained/lost by using this speed up factor.

$\large 100 + 100\times (1 - 0.75) = 100 + 25 = 125$

This is what we are looking for. Lets replace 100 with t and 0.75 with s and doing some algebra:

$\large t + t\times(1-s)$

$\large t\times(1+(1-s))$

$\large (2-s)\times t$

We arrive at the following formula which nicely expresses the relationship between the speed up factor and t:

$\large F(t,s) = (2-s)t$

Here are the assumptions that we are making:

$\large s > 0\\ t > 0$

If s where to equal 0 it would indicate that the video was paused and therefore t should become infinite. This formula doesn’t work that way. Negative values for s would indicate rewinding and that t should be going backwards.

t       s     Actual (s)
956   1.25      717
555   1.05      525.15
1234  0.80      1480.8