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From: AMuzi <am@yellowjersey.org>
Newsgroups: rec.bicycles.tech
Subject: Re: Bicycle physics question
Date: Wed, 19 Jun 2024 19:54:42 -0500
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On 6/19/2024 7:46 PM, Frank Krygowski wrote:
> On 6/19/2024 6:39 PM, bp@www.zefox.net wrote:
>> Jeff Liebermann <jeffl@cruzio.com> wrote:
>>> On Tue, 18 Jun 2024 00:16:01 -0000 (UTC), 
>>> <bp@www.zefox.net> wrote:
>>>
>>>> While out for a motorcycle ride this morning a question
>>>> applicable to both bicycles and motorcycles came to mind:
>>>>
>>>> When a bike/cycle is leaned into a turn, its center of 
>>>> gravity
>>>> is lowered.
>>>
>>> Gravity doesn't move.  However, your center of mass does 
>>> move and is
>>> lowered.
>>> <https://en.wikipedia.org/wiki/Center_of_mass>
>>>
>>>> That would seem to remove some potential energy.
>>>
>>> True, but it's a tiny amount of energy.
>>>
>>> Potential_energy = mass * gravity * height
>>> or
>>> joules = kg * 9.8 meters/sec^2 * meters
>>>
>>> Notice that it's the same change in potential energy 
>>> whether you're
>>> moving of standing still.  You could be riding furiously 
>>> or at a
>>> traffic light, and the change in potential energy would 
>>> be the same.
>>> Your forward motion is also not involved in the potential 
>>> energy
>>> calculation, because it is perpendicular to force vector 
>>> (gravity).
>>>
>>> If you were to lean the bicycle over 1/2 meter and you 
>>> and your
>>> bicycle weigh 80 kg (176 lbs), the change in potential 
>>> energy would
>>> be:
>>> Potential_Energy(change) = 80 * 9.8 * 0.5 = 392 joules or 
>>> 392
>>> watt-seconds
>>>
>>> <https://www.omnicalculator.com/physics/potential-energy>
>>> I like calculators that allow me to mix metric and 
>>> imperialist units.
>>>
>>>> To undo the lean, the wheels have to be steered back under
>>>> the CG, which requires pedal effort on the bicycle and 
>>>> extra
>>>> throttle on the motorcycle.
>>>
>>> Correct.  Assuming 100% efficiency (most of which is lost in
>>> compressing the tires), in the above example, you will 
>>> need to supply
>>> 392 joules of energy to return to an upright position.  
>>> Note that the
>>> energy is supplied only in the upright direction 
>>> (perpendicular to the
>>> ground) and does not involve anything in the forward 
>>> direction.
>>>
>>> There are some interesting comments in this discussion:
>>> <https://www.bikeforums.net/advocacy-safety/288303-what-makes-bike-turn.html>
>>>
>>>> But, leaning a bike/motorcycle doesn't seem to make it go
>>>> perceptibly faster, so if it takes work to stand it back 
>>>> up,
>>>> where did the energy of leaning over go?
>>>
>>> It didn't go anywhere.  It's all POTENTIAL energy, not 
>>> kinetic energy.
>>> You can use potential energy to do work.  Only kinetic 
>>> energy can do work.
>>         ^
>>       can't <-typo?
>> Potential energy can certainly do work, think of a 
>> trebuchet. Potential
>> energy is lost in leaning. Tom thinks it's going into tire 
>> friction,
>> we all seem to agree the amount is smallish compared to 
>> the KE of
>> the bike and dissipation caused by air drag making it hard 
>> to detect..
> 
> I haven't thought deeply about this, but I suspect there's 
> no practical change in potential energy due to the lean. But 
> I'll admit my thinking on this is both somewhat fuzzy and 
> esoteric. Here goes:
> 
> The leaning of the bike+rider occurs only when the bike is 
> turning - that is, undergoing a lateral acceleration. The 
> amount of lean is precisely what's necessary to balance the 
> vector sum of gravity (or its acceleration) and the lateral 
> acceleration.
> 
> One of the concepts that kick-started Einstein toward 
> relativity was the fact that gravity (or its acceleration) 
> and linear acceleration are indistinguishable. Specifically, 
> he realized that no measurement done inside an isolated 
> elevator can tell whether it's gravitational force or upward 
> acceleration that causes a passenger to stay in contact with 
> the floor.
> 
> 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. No change in PE, no resulting 
> change in KE.
> 
> Again, a bit fuzzy. I haven't bothered to work out all the 
> details. But for now, that's my bet.
> 

That all seems right to me, assuming CG means distance to 
tire contact and not height.
-- 
Andrew Muzi
am@yellowjersey.org
Open every day since 1 April, 1971