1.35 horsepower, but I encourage someone to poke holes in my result.
Methodology:
I figure that since there is no actual work being done, all the energy must be dissipated as heat. Given that there is a distance of 17 inches (1.41667 ft) from the wheel to the tensioner set at 20 lbf, it can be known that there is a total torque on the wheel of 28 1/3 lbft.
28 1/3 lbft at 1500 RPM gives a power of 8.1 hp. Given that all energy is dissipated as heat, and that there are six identical, equally placed brake shoes, then each of the six brake shoes dissipates 1.35 horsepower.
Engine Indicators, Power, Tuning

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 fredrosse
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Re: Engine Indicators, Power, Tuning
The common formula for brake horsepower:
BHP = Torque (LBfft) x RPM / 5252
for this case, the torque is Force x Moment Arm = 20 LBf x (6 + 11 )/12 ft = 28.3 LBf  ft
Hence the horsepower is: 28.3 * 1500 / 5252 = 8.1 Brake Horsepower
746 Watts is equivalent to 1 Horsepower, so the total Brake Power in Watts = 8.1 HP x 746 Watts/HP = 6037 Watts
And the Average power dissipated as heat on the brake shoes is 6037 Watts / 6 shoes = 1006 Watts per shoe, (so Tom is correct at average 1.35 HP per shoe) but that is an Average.
The question is asking for the heat dissipated at EACH shoe, and they vary considerably between the six shoes. That is the question which had all the engineers stumped.
BHP = Torque (LBfft) x RPM / 5252
for this case, the torque is Force x Moment Arm = 20 LBf x (6 + 11 )/12 ft = 28.3 LBf  ft
Hence the horsepower is: 28.3 * 1500 / 5252 = 8.1 Brake Horsepower
746 Watts is equivalent to 1 Horsepower, so the total Brake Power in Watts = 8.1 HP x 746 Watts/HP = 6037 Watts
And the Average power dissipated as heat on the brake shoes is 6037 Watts / 6 shoes = 1006 Watts per shoe, (so Tom is correct at average 1.35 HP per shoe) but that is an Average.
The question is asking for the heat dissipated at EACH shoe, and they vary considerably between the six shoes. That is the question which had all the engineers stumped.
 Lopez Mike
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Re: Engine Indicators, Power, Tuning
Wouldn't the correct arm length be 18.6832"? That's the distance from the point of measurement to the center of the axle.
The last time I looked the standard h.p. was 33,000 ft/lb per minute or 550 ft/lbs per second.
I get an arm length of 1.557 feet with a force of 20 lbs. which gives me a torque of 31.14 ft/lbs.
1500 rpm times 31.14 lbs equals 46710 ft/lb per minute which is 1.42 h.p.
divided by six gives a little less than .236 h.p per block or 176 watts per block.
The fussing over whether all six blocks get equal torque I think is a red herring. I think they share it equally but I could be persuaded otherwise.
Let the bricks start flying!
The last time I looked the standard h.p. was 33,000 ft/lb per minute or 550 ft/lbs per second.
I get an arm length of 1.557 feet with a force of 20 lbs. which gives me a torque of 31.14 ft/lbs.
1500 rpm times 31.14 lbs equals 46710 ft/lb per minute which is 1.42 h.p.
divided by six gives a little less than .236 h.p per block or 176 watts per block.
The fussing over whether all six blocks get equal torque I think is a red herring. I think they share it equally but I could be persuaded otherwise.
Let the bricks start flying!
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Einstein  Extraordinary mind
Me  Never mind.
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Einstein  Extraordinary mind
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 fredrosse
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Re: Engine Indicators, Power, Tuning
Wouldn't the correct arm length be 18.6832"? That's the distance from the point of measurement to the center of the axle.
ANS TORQUE IS FORCE X THE MOMENT ARM TAKEN AT RIGHT ANGLE TO THE FORCE VECTOR, HENCE 6 + 11 = 17 INCHES.
The last time I looked the standard h.p. was 33,000 ft/lb per minute or 550 ft/lbs per second.
ANS THAT IS CORRECT
I get an arm length of 1.557 feet with a force of 20 lbs. which gives me a torque of 31.14 ft/lbs.
1500 rpm times 31.14 lbs equals 46710 ft/lb per minute which is 1.42 h.p.
ANS IN A ROTATING SHAFT, THE WORK IS TORQUE X THE DISTANCE AROUND A CIRCUMFERENCE, MULTIPLY X 2 pI
divided by six gives a little less than .236 h.p per block or 176 watts per block.
The fussing over whether all six blocks get equal torque I think is a red herring. I think they share it equally but I could be persuaded otherwise.
ANS THE THIN STEEL BAND HAS DIFFERENT ANGLES FORCING THE BLOCKS AGAINST THE WHEEL, SOMETIMES THE BAND PULLS PARALLEL TO THE WHEEL ON ONE SIDE, THUS NO FORCE INTO THE WHEEL, AND OTHER BANDS ORIENT TO PULL THE SHOE INTO THE WHEEL ON THE OTHER SIDE, DIFFERENT GEOMETRIES AROUND THE SHOES.
Let the bricks start flying!
ANS TORQUE IS FORCE X THE MOMENT ARM TAKEN AT RIGHT ANGLE TO THE FORCE VECTOR, HENCE 6 + 11 = 17 INCHES.
The last time I looked the standard h.p. was 33,000 ft/lb per minute or 550 ft/lbs per second.
ANS THAT IS CORRECT
I get an arm length of 1.557 feet with a force of 20 lbs. which gives me a torque of 31.14 ft/lbs.
1500 rpm times 31.14 lbs equals 46710 ft/lb per minute which is 1.42 h.p.
ANS IN A ROTATING SHAFT, THE WORK IS TORQUE X THE DISTANCE AROUND A CIRCUMFERENCE, MULTIPLY X 2 pI
divided by six gives a little less than .236 h.p per block or 176 watts per block.
The fussing over whether all six blocks get equal torque I think is a red herring. I think they share it equally but I could be persuaded otherwise.
ANS THE THIN STEEL BAND HAS DIFFERENT ANGLES FORCING THE BLOCKS AGAINST THE WHEEL, SOMETIMES THE BAND PULLS PARALLEL TO THE WHEEL ON ONE SIDE, THUS NO FORCE INTO THE WHEEL, AND OTHER BANDS ORIENT TO PULL THE SHOE INTO THE WHEEL ON THE OTHER SIDE, DIFFERENT GEOMETRIES AROUND THE SHOES.
Let the bricks start flying!

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Re: Engine Indicators, Power, Tuning
Not to mention that the friction on in the blocks will change the amount of tension in the belt, in turn changing which blocks are pulled hardest into the wheel.fredrosse wrote: ↑Mon Sep 03, 2018 9:39 pmThe fussing over whether all six blocks get equal torque I think is a red herring. I think they share it equally but I could be persuaded otherwise.
ANS THE THIN STEEL BAND HAS DIFFERENT ANGLES FORCING THE BLOCKS AGAINST THE WHEEL, SOMETIMES THE BAND PULLS PARALLEL TO THE WHEEL ON ONE SIDE, THUS NO FORCE INTO THE WHEEL, AND OTHER BANDS ORIENT TO PULL THE SHOE INTO THE WHEEL ON THE OTHER SIDE, DIFFERENT GEOMETRIES AROUND THE SHOES.
Let the bricks start flying!
I'm working on a script to calculate the steady state condition, but in real life I'd probably just calculate the total heat dissipation, then use an infrared thermometer to determine which blocks were hottest, and using that assign values based on their temperatures over ambient.
 fredrosse
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Re: Engine Indicators, Power, Tuning
Tom is correct again, tension in the metal band changes between shoe blocks, another complication. With an average heat value of a little more than 6000 Watts of power on six shoes, and some shoes getting considerably more than 1000 watts, the wood shoes would start smoking pretty good in a short period of time. That is why I had to keep the Prony brake flooded with water to keep from burning the thing up.