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From: Catrike Ryder <Soloman@old.bikers.org>
Newsgroups: rec.bicycles.tech
Subject: Re: Extensive article on Rivendell and Grant Petersen
Date: Wed, 25 Sep 2024 11:45:46 -0400
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On Wed, 25 Sep 2024 11:01:17 -0400, Frank Krygowski
<frkrygow@sbcglobal.net> wrote:

>On 9/25/2024 5:39 AM, Catrike Ryder wrote:
>>  
>> A few weeks ago, after posting about braking, I tested the Catrike's
>> brakes at 15 MPH. I stopped at about 6 feet, keeping the chain rings
>> off the ground.
>
>No you didn't, unless your "about 6 feet" has a tolerance of something 
>like 50%.
>
>For the engineers in the crowd: It's a simple constant (negative) 
>acceleration problem. Acceleration (or deceleration) is given by V^2/2X 
>where V is initial speed, X is stopping distance. 15 mph = 22 ft/s
>
>(22 ft/s)^2/(2*6ft)= 40.33 ft/s^2 deceleration. That's 1.25 times the 
>acceleration of gravity. For that, you'd need tires with a coefficient 
>of friction of at least 1.25, which would be very, very unusual. (0.9 is 
>a typical upper limit.) But more important, you'd need to _immediately_ 
>apply the brakes to the very limit of traction with no skidding; and 
>you'd need no weight on the unbraked rear wheel, so all the decelerating 
>mass was contributing to braking traction. You'd also need exactly the 
>same amount of braking on each front wheel so as to prevent a spin, 
>given that the rear wheel would have to be raised.
>
>Oh, and whether or not the rear wheel would raise to put all the weight 
>into front wheel traction depends on the geometry of the bike+rider. The 
>elevation angle of the total center of mass would have to be precisely 
>right, not too high nor too low.
>
>All this is based on the physics of the real world. Those living in 
>other universes should post their math, or their videos.

Yawn...

--
C'est bon
Soloman