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
```