Cody,

The torque and horsepower curves cross at 5252 RPM.

(Only in English units)

In other words, each Lb-foot of torque gives you 1 HP at 5252 RPM.

You and I know that engine torque curves are not quite flat, but to make rough estimates we can act as though they are. Good enough for back-yard work.

HP is proportional to RPM, in other words half the RPM gives you half the HP.

So an engine that is factory rated to develop 6 HP at 3600 RPM is, by definition, developing (5252/3600) more torque than the engine next to it that develops the same 6 HP at 5252 RPM. And side-note : you probably noticed that a well-tuned engine (no blower, no nitrous, just naturally aspirated) develops about 1 Lb-Foot of torque per cubic inch. Just an accident of James Watt’s choice of strong horses for his horse power measurements. A happy accident that makes a part of this easy to remember. Ball bearings and chain drives are very efficient, so you can call it 100% for rough figures or 95% if you want to get technical. So an engine rated at 6 HP at 3600 RPM is developing 6X(5252/3600) Lb-Ft of torque at 3600 RPM. That is 8.75 LbFT of torque. Now, 61 cubic inches = 1000cc and torque is about equal to cubic inches in the best engines, but this is not a crotch rocket engine, so figure 80% of that scenario, which means divide 8.75 by 0.80 which gives you 11 cubic inches for an engine displacement estimate, and dividing that by 61 cubic inches works out to .179 part of 1000cc, which is .179 liter or 179cc. So your engine is about 180cc. Its torque at max HP is 8.75 LbFt, and you know that max HP occurs to the right of max torque on the power curve, so assuming a flat torque curve pinned at the max power point is not bad for rough figures. It understates peak torque and it is correct one more time in the mid-range. Taking 95% of 8.75 LbFt you get 8.3 LbFt from the little engine that could. Now you just multiply by gear ratios as you know to get your steady-state output shaft torque. This works out great if you are turning a boat prop. But the chunker has energy stored in a flywheel. The amount of force the shark fin will apply to the wood is the output shaft torque divided by the lever-arm in feet. It is from the center of the shaft to the center of pressure where the shark fin is crushing wood fibers. Your double reduction 11:60 gear train is better than you thought. It’s 29.75:1 because the ratios multiply on each other; they are not additive. Think about a double set of 3:1 pulleys. If you turn the input shaft 9 times, the jack-shaft will turn 3 times, and the output sprocket will turn 1 time, so it works out as 3 X 3 = 9:1 overall gear ratio. So with your serpentine belt primary drive reduction of 8:1 multiplied by your double reduction chain drive, the output torque from the little 6 HP engine will be (29.75 x 8) x 8.3 LbFt = 1975 LbFt so round it to 2000 LbFt average. And the output speed will be 3600 / 238 = 15 RPM or 1 rev every 4 seconds. OK, maybe a tad slow. Now, let’s use backyard math which is good enough for our purposes. We are committed to increasing the flywheel mass until the chunker does not get stuck, right? It’s fair to estimate that 1/2 of the revolution of the output shaft is the portion in which serious force is being exerted by the shark’s tooth. So 4000 LbFt will be exerted on the keyway in the 1 inch shaft. I’ll assume the shaft is made of SAE 1045 which has a yield strength of 75,000 Lbs per square inch, and the key is equally strong. Eyeballing that hub, it looks like it’s 1 3/4 inches long. A standard 1/4" key engages 1/8" so the area of the side of the keyway which is taking all the force is .125x1.75=.219 inches squared. The key’s effective lever arm is 7/16" or .03646 foot. 4000 LbFt / .03646 foot = 109,710 pounds exerted on .219 square inches of area, so the stress is 510000 PSI, which is 6.5 times more stress than the keyway can handle. The keyway can belt out .219 InSq X 75,000Lb/InSq = 16,425 Lbs of tangential force at a lever arm of .03646 Ft. = 600 LbFt of torque. The shaft itself is good for 680 LbFt. 600 LbFt / .42 Ft = 1429 Lbs distributed over 1 InSq of shark tooth edge smashing into 4 inch diameter wood = 1429 PSI available to crush the wood which is adequate only for soft wood. Sharpening the shark’s tooth will help some, but after some penetration, the full width of the tooth is bearing upon the wood. This chunker will be adequate for 4 inch softwood and 2 inch hardwood. Just don’t let some jerk stuff a 4 inch oak tree branch into it, calling on the 4,000 LbFt momentarily available, bending your ouput shaft or shearing its key. I suggest you design it with two of the keyed hubs and two bearings carrying the shaft, one on each side of the shark tooth, so the shaft will not be cantilevered. Now, the electric motor option. I like the foot-pedal switch idea for safety. It’s probably a 3600 RPM motor (3450 RPM actual speed, but close enough). So the 15 RPM output speed will be the same (14.4 RPM for the typical know-it-all uninvited neighbor, haha) The torque will be 1/3 nominally, but electric motors increase their torque in response to load, so the loss is not as bad as the numbers look. But expect to have to wait for it to recover RPMs or put smaller pieces of wood in it. To some extent the motor armature is a flywheel, as long as the primary drive belt does not slip. Overall, it will work but it won’t be mega-strong like Wayne’s, eating slab sandwiches! I agree with your idea to try it with minimal flywheel mass at first. It will be better to have it stall on oversize wood than to keep shearing the keys. I’ve tried hard to be kind to your idea here, but the output shaft really needs to have a larger diameter. -Mark the RedOak guy