2016-03-12

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
```