Author Topic: How important is weight?  (Read 3669 times)

CrazyEnglishTriathlete

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Re: How important is weight?
« Reply #25 on: December 02, 2011, 10:52:58 pm »
Weight is not a problem on the flat and a slight advantage on downhills.  Its a big disadvantage on steep uphills.

However, this still doesn't explain why every November somebody goes round and cranks up all the hills to make them steeper.
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Re: How important is weight?
« Reply #26 on: December 04, 2011, 10:02:38 pm »
One of the things that got me thinking was a comment by a PBP rider. He described his 'bent as "eye wateringly heavy" once loaded and yet 900 kms into the ride was averaging 17mph in a particularly hilly section and passing DF riders. Weight obviously does matter but the evidence seems to suggest it is not always the most important factor.

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John

Re: How important is weight?
« Reply #27 on: December 04, 2011, 10:30:08 pm »
Ask some of those velomobile riders - usually an easy 30kg of machine, and yet tearing along at incredible average speeds.
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Wothill

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Re: How important is weight?
« Reply #28 on: December 05, 2011, 09:40:19 pm »

Wind resistance goes up at square of the speed.
Power needed to go at that speed goes up at the cube.

Hang on. I agree with the first: wind resistance is proportional to the square of the speed, but the second surely contradicts that. How could you measure wind resistance other than in units of power? Or did I miss a step in the logic?


Wind resistance is a force - you can measure it in lbs or Kg, but purists would use Newtons.
To get the power required to maintain that force you need to bring time and distance into it, as there's clearly no way that you can sustain the speed that you just calculated the wind resistance for if the force you can generate is at a lower speed (the force you can generate at the driven wheel is a product of engine torque (rider strength) x gear ratio).  Sheldon Brown's "gain ratio" is useful for working that out, since it can directly translate leg strength to tractive effort.
You have to multiply by the speed again to get the power figure from the torque, thus increasing from the square to the cube.
Sorry. You are quite right of course. I have tended to think of wind resistance in a rather woolly way as the power required to overcome it but as you point out it that proportional to the cube of the speed through the air - not the square. Since most bents are much more aerodynamic than uprights you would expect them to be faster, by comparison, than they are. I suppose this is another indication of the apparent inability to put so much power through the pedals as on an upright.