Jake Anderson wrote:
> brembo 6 piston caliper, 55 square centimeters worth of piston area
> decent brake pad CF is ~.8
> push the pressure to 3000PSI say.
>
> a 10" blade is say .1" thick
> weight is ~1kg, KE is 1500J
>
> converting to a linear motion to make life simple
> KE = .5MV^2
> sqrt(KE /.5M) = V
>
> V=54m/s
> ~=200km/h which seems to be ~ the tip speed of the blade as calculated
> by others so it seems good.
>
> clamping force on the disk is 8.5 square inches * 3000PSI = 25500
> pounds = 11590kg = 113590N
> deceleration force = .8 * clamping = 90872N
> acceleration = F/M = A = 90872M/s/s (~9000G)
> v^2/2a = s
> 57^2/2*90872 = 0.017876794 meters
> so about 20mm give or take.
>
> seems fast enough for me?

But you're leaving out lots of stuff.  What mechanism is going go from 0 to
3000 PSI on 8.5 square inches of brake surface, and how long will that take?
The actual braking doesn't sound like the hard part, but actuating the brake
in a very short time is.  The larger you make the brake, that harder that
will be.


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