Re: Bicycle physics question

On 6/20/2024 10:17 AM, Frank Krygowski wrote:

On 6/20/2024 5:05 AM, zen cycle wrote:

On 6/19/2024 8:46 PM, Frank Krygowski wrote:

>
So it's logical to treat as identical sorts of vectors both the upward force on a cyclist (fighting gravity) and the lateral force on a turning cyclist (pushing him into a curve). It's the resultant of those two forces against which potential energy is determined. And again, the cyclist's angle is always what's necessary to exactly balance that resultant.
>
So in that reference frame, there is no reduction in potential energy. The bike+rider's CG is always the same distance above the tire.

>
REally? This CG:
>
https://jacoblund.com/cdn/shop/products/f60bc56d059f05627a4511dcb9495818_1500x.jpg?v=1679657105
>
is the same as this CG?
>
https://cycling.today/wp-content/uploads/2017/05/Peter-Sagan-min.jpg
>
>
No

 That's a different problem - no lateral acceleration - but it's valuable for thinking about Bob's original question.
 Yes, lowering one's CG that way decreases potential energy. But the act of lowering does not cause an acceleration.
 If a cyclist were to repeatedly stand tall on the pedals, then crouch as low as possible, then stand back up, he would not experience accelerations and decelerations each time he did that.
 That case is easier to explain. The relevant vectors are perpendicular to each other.
 

Speaking of different aspects, a full upright stand on a descent and then folding into an aero position makes for a dramatic speed change.
--
Andrew Muzi
am@yellowjersey.org
Open every day since 1 April, 1971