Timeline for Which type of course is the fastest for a 10 km TT? A completely flat course or a course with 5 km uphill and 5 km downhill?
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| Oct 23, 2020 at 10:21 | comment | added | Vladimir F Героям слава | @leftaroundabout No, I do not buy that at all. If the start and the end is at the same height you will spend a lot of time climbing back to the starting height. You can only use brachistochrone for a descent and of course you can indeed always optimize the descent curve, including the drag with some approximations and everything, it is simple mechanics, but then you have tu climb up and it will take a lot of time. And of course, if you had a flat tarmac instead of a BMX pumptrack, you would be MUCH faster. | |
| Oct 23, 2020 at 10:17 | comment | added | leftaroundabout | @VladimirF well, it doesn't apply because it's impossible to have a full course remotely resembling a brachistochrone (let alone one without air drag). But if you could have such a course, then it would indeed give the fastest time for a cyclist: the descend alone readily results in speeds that make the bit you can build up by pedalling for the same amount of time quite insignificant. — Where the descent-ascend-advantage principle is very evident though, albeit on a much smaller scale, is in the skatepark and on BMX tracks. | |
| Oct 23, 2020 at 9:52 | comment | added | Vladimir F Героям слава | @leftaroundabout The good old brachistochrone problem does not really apply here in any way. That is a trajectory for a fastest free fall on a curve between two points. We have active movement against friction here, or speed on the flat is positive and finite. | |
| Oct 23, 2020 at 9:38 | comment | added | leftaroundabout | @VladimirF true, but without any ærodynamic drag the optimal course is not flat but it first decents, then rises again. The good old brachistochrone problem... | |
| Oct 21, 2020 at 13:55 | history | edited | Adam Rice | CC BY-SA 4.0 |
typo fix
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| Oct 21, 2020 at 10:22 | comment | added | Vladimir F Героям слава | @ChrisH I see, yes. | |
| Oct 21, 2020 at 9:55 | comment | added | Chris H | @VladimirF yes, but this answer makes (IMO) too close a comparison to the effects of headwind, unlike yours | |
| Oct 21, 2020 at 9:53 | comment | added | Vladimir F Героям слава | @ChrisH Really, even without any aerodynamic drag at all in vacuum you still lose time. It is just the simple way how speed averaging works. | |
| Oct 21, 2020 at 9:25 | comment | added | Chris H | Yes and no - the headwind/tailwind factor is because the power lost to air drag scales with the cube of speed, but climbing power scales linearly with speed. However at constant effort this same effect means the effective headwind (even in still air) is greater on the downhill segment, and the reduction on the uphill doesn't make up for it | |
| Oct 21, 2020 at 8:14 | comment | added | ojs | Well, part of the question was "is one of these courses faster than the other" | |
| Oct 21, 2020 at 7:57 | comment | added | Vladimir F Героям слава | @Michael As in any other race, I do not see a point. | |
| Oct 21, 2020 at 7:56 | comment | added | Michael | @VladimirF: You lose time, but if the others lose more time you still win the race. | |
| Oct 21, 2020 at 7:54 | comment | added | Vladimir F Героям слава | The problem with this argument is that even if you do get all the advantage from the 5 km/h wind and go 25 km/h one way and 35 km/h the other way, you still lose (yes, meaning lose time). It is not necessary to have any losses whatsoever. It is more obvious when it is 10 km/h one way and 50 km/h the other way. | |
| Oct 21, 2020 at 1:19 | history | answered | Adam Rice | CC BY-SA 4.0 |