# 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

• Delta D divided by delta V. Oct 13, 2013 at 22:10
• This question appears to be off-topic because it is about trolling Oct 14, 2013 at 1:36
• Speaking of trolling: @DanielRHicks don't the Deltas cancel? Oct 14, 2013 at 2:35
• @andy256 - "Delta" means "difference". Oct 14, 2013 at 10:53
• @ChrisW - Since he wants to do the math "on the fly" I'm guessing it's about fly fishing, not trolling. Oct 14, 2013 at 19:41

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.

• Yeah, basically delta D divided by delta V. Oct 14, 2013 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, 2013 at 6:55
• so it's dd/d(dx/dt)? Oct 16, 2013 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) Oct 16, 2013 at 9:17