On 13/07/10 22:40, Olin Lathrop wrote:
> 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.
>    
As I said in my original post a hydraulic acumulator connected to shop 
air would provide you with a 3000PSI reservoir, then a valve analogous 
to a fuel injector or ABS valve to apply that pressure to the cylinders, 
the flow should be minimal so the pressure ramp time should be as well. 
Besides you have complete control over these aspects of it so its "just 
a matter of engineering". The basic principle seems sound, provided you 
don't fling the carbide off the blade it seems a valid solution.
Any system that is going to pull the blade out of the way of fingers is 
going to need to move at a similar pace so the carbides falling off is 
going to be the make or "brake" part of the problem.

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