# What will be the time to close a gap to catch a breakaway? [closed]

how much time is required to close a gap of 100 metres, in a cycling race, for different differences in speed of leading and chasing group.

i need a tool to do the math on the fly. so that if one knows what speed the chasing group, one should be able to decide the speed to keep gap constant

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## closed as unclear what you're asking by ChrisW, amcnabb, freiheit♦Oct 27 '13 at 5:38

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Delta D divided by delta V. – Daniel R Hicks Oct 13 '13 at 22:10
This question appears to be off-topic because it is about trolling – ChrisW Oct 14 '13 at 1:36
Speaking of trolling: @DanielRHicks don't the Deltas cancel? – andy256 Oct 14 '13 at 2:35
@andy256 - "Delta" means "difference". – Daniel R Hicks Oct 14 '13 at 10:53
@ChrisW - Since he wants to do the math "on the fly" I'm guessing it's about fly fishing, not trolling. – Daniel R Hicks Oct 14 '13 at 19:41

## 1 Answer

how much time is required to close a gap of 100 metres, in a cycling race, for different differences in speed of leading and chasing group

For a speed difference of 1 metre per second, the time is 100 seconds.

For a speed difference of 2 metres per second, the time is 50 seconds.

For a speed difference of 0.5 metres per second, the time is 200 seconds.

Generally, for a speed difference of "`x`" metres per second, the time is "`100 divided by x`" seconds.

i need a tool to do the math on the fly. so that if one knows what speed the chasing group, one should be able to decide the speed to keep gap constant

To keep the gap constant, the lead group must maintain exactly the same speed as the chasing group.

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Yeah, basically delta D divided by delta V. – Daniel R Hicks Oct 14 '13 at 19:42
If you know the speed difference in kph only, the "closing time" for 100m in seconds would be 360 divided by the speed difference in kph. But yeah, dd/dv. – arne Oct 15 '13 at 6:55
so it's dd/d(dx/dt)? – imel96 Oct 16 '13 at 6:07
It's ((x<sub>1</sub> - x<sub>2</sub>) / (v<sub>2</sub> - v<sub>1</sub>)) if v<sub>2</sub> and v<sub>1</sub> are on the same path, except when (v2 = v1) – ChrisW Oct 16 '13 at 9:17