Wanted to start this thread talking about the required calculations for the shredder.
The calculation for torque is from the shear stress required to cut a specific material. The shredder uses a mode of shear to cut material which is a perpendicular force applied to a material. In this case 2 perpendicular forces that are equal and opposite in direction. the 2 perpendicular forces are parallel to each other but opposite in direction.
Shear is calculated from a rule of thumb percentage of the yield stress(its something like 80% of yield stress) of the material you are trying to cut. (google yield stress of your material). stress is the Force applied divided by the Area that the force is applied over =F/A. The Area is found by the cross section area of the material of cut that the shredder blade contacts. Then to determine the power of the motor required is from the torque the motor puts out at a given rpm. Torque is Force times Distance F*D. So to solve for force at the cutting area you will divide the torque by the RADIAL(Radius of cutter) distance to the cutting TIP (worst case) . This solves to (F*D)/D = F. Now Equation to find horsepower is RPM (rotation per minute) * Torque all divided by 5252
So lets say the contact cross section area(A) is calculated from a Triangle sweep of shape of cut
(2.5″ x .4″)/2″ = .5″^2 (the worse case from design of my shredder)
Radial distance to tip is 3.5″ Diameter/2 (my design of shredder) R = 1.75″
I googled Yield Strength of HDPE and came up with 4350 PSI (F/A aka pressure).
So i am going to take 80% of that for rule of thumb i get 3500 PSI.
So 3500 PSI = F/A.
I solve for Force so Force = 3500 * .5 = 1750 That equals 1750 Lbs of force.
Now solve for how much torque i need from my motor.
Torque = Force * Distance so Torque = 1750 * 1.75″ = 3063 inch-pounds of torque.
lets say i have a motor that rotates at 5 rpm.
For horsepower (3063 * 5) = 15315 in-lbs/sec / 12 = 1276.25/ 550 =2.32HP
Then we have something called S.F. (safety factor) which is basically the needed power multiplied by the Safety Factor which is basically a allowance for unaccounted variables or harder materials that are trying to be shredded etc. in this case lets say S.F. of 2 so 2.32 * 2 = 4.6HP.
The reason why this is so high is because we are cutting the max cross section of HDPE that my shredder can handle.
So what if it is human powered. how long of a bar “moment arm” would you need to shred that max cross section area. so we need 3,063 inch-lb of torque. i weigh 160 lbs. so take 3063 and divide by 160 lbs to get a 19.14 inch long bar witch is almost 2 feet which is very manageable. Meaning that the machine will be way cheaper and easier to power if it was just a long bar and you hang on it.
So when you do your calculations plug in the design of your shredder and use the yield stress of the HARDEST material you are trying to cut. Then find your HP Minimum then multiply it by at least 2 to allow for some other variables to get the motor you need.
Motor selection is very complex topic because you have to take account motor efficiency at different rpm and torque ranges. Brushless motors are most efficient at 80-90% their max RPM a efficiency of 80-90%. However at that RPM range torque is at 1/8 its maximum output torque. Luckily when a Motor is rated a specific HP rating that rating is at its maximum efficiency. Maximum efficiency of the motor is found at approximately half its MAXIMUM output HP. So that means when you load it even more the RPMs will drop to compensate for the required torque needed.
That is where gearboxes come in. The purpose of gearboxes is to try to have your cake and eat it too. In other words to run your motor at its max efficiency RPM and horsepower rating but “gear down” the rpm so that you also get the desired torque you want. So you go to your gearbox to trade RPM for torque. To do this you have a ratio such as 10:1 which means that your input gear is 10 times smaller than the output gear. Meaning the input gear has to turn 10 times to turn the output gear once. This cuts down your rpm by 10 times but increases your torque 10 times (not exactly). The higher the ratio the lower the output RPM but the higher the torque. And the reason why i said not exactly above is that gearboxes are not 100% efficient and you will lose power as heat from friction from the gears meshing (most gearboxes are around 90% efficient). So for a shredder the main question you have to ask yourself is mainly how low of RPMs can i live with that will determine your motor and gearbox selection.
Lets say we have a 1.5KW(~2 HP) Spindle motor for engraver/router. Reading off this spec sheet
rated Power: 1.5KW(~2HP)
Rated Speed: 1,000 RPM
Max RPM: 2400 RPM
So what is our torque. (RPM * Torque)/5252= HP
We have Rated RPM and Rated Power
Solve for Torque = (HP * 5252) / RPM = 10.5 Ft-lbs (pounds)
Alright so gearbox solve now
Lets say we are ok with 5 RPM for output
so from 1,000 RPM to 5
1,000/5= 200 so we would have a 200:1 ratio
So what is output torque
10.5 * 200 = 2100 Ft-lbs
Massive Torque. Low RPM. Easy Shredding. Or might tear machine apart.
Another thing you can do is run your motor at like 90% its maximum HP (dont want to run at 100% you will burn motor up) rating that will give you even more torque. However you will lose a lot of energy as heat as your motor will be running at 60% energy efficiency rather than 80% meaning you are spending 20% more energy than you need to.