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From: AMuzi <am@yellowjersey.org>
Newsgroups: rec.bicycles.tech
Subject: Re: Bicycle physics question
Date: Thu, 20 Jun 2024 10:27:57 -0500
Organization: Yellow Jersey, Ltd.
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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