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I'm building a quadricycle. This sociable is planned to be around 120 lbs / 54.5 kg. The only "Guide" to proper rotor size versus weight is one I found from Magura. Average pilot/rider weight I figure to be around 200 lbs / 90.7 kg. This would add up to 520 lbs / 236 kg. Cut in half to use the guide sheet, (two wheels versus four), and I get total weight of 260 lbs / 118 kg. Street or off-road use says 160mm rotors all around. The other consideration I have to take in is, recumbents have most of the weight riding in the back, 60/40. During braking there is a weight shift I know, but does it go over 50/50 during hard braking? 160mm rotors just don't sound right/enough. And I am using duplex dual cable levers, one front, one rear.

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    Welcome to Bicycles @SRHampton. As with all new members, we recommend taking the tour to make best use of the site, and since you're asking, How to Ask is worthwhile also. I guess you know who the riders are going to be, but 90 kg each seems on the high side. However, it's better to make pessimistic design assumptions. But what about using 180mm rotors? They seem to be cheaper ATM. – andy256 Sep 9 '16 at 7:28
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  • Most commercial quadricycles are inherently slow (single speed and not even geared high). As the energy you'll need to dissipate scales with the quare of the speed but only linearly with the mass, you may not need as much braking as e.g. a downhill bike. However if you're building for speed you may have issues. Commercially drum brakes on the drive axle seem common – Chris H Sep 9 '16 at 10:50
  • To Chris H 21, My design includes a drive system to fill the gap between automobiles, and bicycles or velomobiles. I am shooting for an average speed of 25 mph, with a maximum of around 35 mph. – S R Hampton Sep 10 '16 at 8:29
  • To andy256, 180mm seems to be quite common, and there are many types available. Thanks. Plus I will be checking out the site over the weekend! – S R Hampton Sep 10 '16 at 8:30
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You have two considerations, energy dissipation(will they overheat) and braking force(will you stop in a short distance or even stop at all on a steep down hill). I doubt overheating will be an issue, but having 4 brakes will divide the force you apply to the levers. Using hydraulic calipers will overcome the force division issue but may encounter a fluid volume(lever stroke) issue.

The calculations needed are newtons of force times meters divided by seconds equals watts.
Maximum braking force is limited by the tire-road coefficient of friction which is less than 1.0 in common practice, 0.85 is normal, above one is only common in specially selected warm race car tires. So, mass times 9.8(converts Kg mass to newtons under earth gravitational acceleration) times 0.85 gives maximum braking force in newtons. Newtons divided by mass in Kg gives acceleration in meters per second per second. Braking is acceleration, just in the other direction.
So if you have 520kg the maximum braking force is 4332N, acceleration of 8.33m/s^2. From 15m/s to zero = 1.8 seconds (on level ground) Kinetic energy in joules is 0.5 times Kg times velocity squared. In this case 58500j. Divided by 1.8 seconds is 32.5Kw divided by 4 brakes is 8.1Kw per rotor. Heat capacity of steel ranges from 460-510 j per k, 58500/480=122k temp rise in one kg of steel. 160mm rotors seem to range 50-140g, 100g is common so 400g total, 58500/(480*0.4)=305k increase, this is acceptable, it also doesn't take into account the heat loss to radiation and the air during the 1.8 seconds.

I don't have the formula for continuous heat dissipation of the rotor at hand, this or testing would be needed for long downhill sections or rapid repeated stops. The watts of heat on a long downhill is zero at terminal velocity and zero at a stand still, peak brake wattage occures at 60% of terminal velocity. The range of 42%-72% T.V. is greater than 0.9 peak power, 0.5 peak power occures at 20% and 88% of terminal velocity. And of course the final impulse to stop at the bottom of the long hill is far less at 20% than at 88% t.v. so even though the rotors should be equally hot just before attempting to stop from either speed they certainly won't be equal after the stop.

Joules of potential energy from a hill is equal to vertical height in meters times gravitational force in newtons. (kg*9.8 here on earth) Joules per second equals watts.

Sorry I'm out of time so I can't give hill slope to braking force equation. Its simple vectors, should be easy to look up.

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