J
Jim Smith
Guest
[email protected] writes:
>
> I'm not sure what is subtracted from which and how it's
> calculated.
>
> 214 pedal Watts for 22.73 mph (10.61 mph) 0% grade
Why 22.73mph? I thought you were interested in 20mph?
> 916 coast Watts for 39.82 mph (17.80 m/s) 7% grade
> 75 kg Mass of bike and rider
> -0.07 slope of 7%
> 10.61 Braked descending speed 23.73 mph = 10.61 m/s
> 17.80 Unbraked descending speed 39.82 mph = 17.80 m/s
>
> 75 * 9.8 * 0.07 * 17.8 = 915.8
>
> (That's the 916 watts already given by the calculator.)
You want to calculate this at the speed you are interested in, not at
17.8 m/s
At 22.73mph (10.61m/s) decending a 7% grade, gravitational potential
energy is changing at a rate of: 75 * 9.8 * 0.07 * 10.61 = 546
jules/second (watts) This is the power input to our bike. Of these
546 watts, 214 are going to overcoming wind and rolling resistance,
leaving 546 - 214 = 332 watts that need to go somewhere so we don't
speed up. You can either use these 332 watts to try and melt the glue
holding on your tubular tires by pushing rubber pads against the rim,
or you can use them to convert donuts into sexy man-flesh by resisting
on your fixie. The choice is yours.
> Subtract from this 916 watts the 214 watts needed to go 20
> mph on the flat, and the rider must provide about 700 watts
> of braking power, which was what I was asking about.
>
> Are we saying that a rider must provide a steady 700 watts
> of reverse pedal power to descend a 7% grade at around 23
> mph instead of around 40 mph?
>
> Or is that 700 watts just what's needed to slow a 40 mph
> descent down to 23 mph, after which a lower reverse pedal
> effort will serve to maintain a steady 23 mph? If so, what's
> the steady-state power needed to stop the bicycle from
> accelerating back up from 23 mph to 40 mph?
>
> Of course, I may be missing something and using the wrong
> terms, but I think that I followed your suggested equation.
>
> I'm not sure what is subtracted from which and how it's
> calculated.
>
> 214 pedal Watts for 22.73 mph (10.61 mph) 0% grade
Why 22.73mph? I thought you were interested in 20mph?
> 916 coast Watts for 39.82 mph (17.80 m/s) 7% grade
> 75 kg Mass of bike and rider
> -0.07 slope of 7%
> 10.61 Braked descending speed 23.73 mph = 10.61 m/s
> 17.80 Unbraked descending speed 39.82 mph = 17.80 m/s
>
> 75 * 9.8 * 0.07 * 17.8 = 915.8
>
> (That's the 916 watts already given by the calculator.)
You want to calculate this at the speed you are interested in, not at
17.8 m/s
At 22.73mph (10.61m/s) decending a 7% grade, gravitational potential
energy is changing at a rate of: 75 * 9.8 * 0.07 * 10.61 = 546
jules/second (watts) This is the power input to our bike. Of these
546 watts, 214 are going to overcoming wind and rolling resistance,
leaving 546 - 214 = 332 watts that need to go somewhere so we don't
speed up. You can either use these 332 watts to try and melt the glue
holding on your tubular tires by pushing rubber pads against the rim,
or you can use them to convert donuts into sexy man-flesh by resisting
on your fixie. The choice is yours.
> Subtract from this 916 watts the 214 watts needed to go 20
> mph on the flat, and the rider must provide about 700 watts
> of braking power, which was what I was asking about.
>
> Are we saying that a rider must provide a steady 700 watts
> of reverse pedal power to descend a 7% grade at around 23
> mph instead of around 40 mph?
>
> Or is that 700 watts just what's needed to slow a 40 mph
> descent down to 23 mph, after which a lower reverse pedal
> effort will serve to maintain a steady 23 mph? If so, what's
> the steady-state power needed to stop the bicycle from
> accelerating back up from 23 mph to 40 mph?
>
> Of course, I may be missing something and using the wrong
> terms, but I think that I followed your suggested equation.
