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From: Frank Krygowski <frkrygow@sbcglobal.net>
Newsgroups: rec.bicycles.tech
Subject: Re: Extensive article on Rivendell and Grant Petersen
Date: Wed, 25 Sep 2024 11:01:17 -0400
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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.


-- 
- Frank Krygowski