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Your input is greatly appreciated however you are mistaken about the portion in bold. I plan to use a digital photointerrupter just like the PM-F25 you linked to. Also i think you might be misunderstanding my explanation about the disks being close together.OK I think I am following...
So the solutions I have written about above are extremely close to the manner of measurement in video #1 in post #1. You'd need to track the positions of two encoders (zeroed out under zero load), query them for their current position simultaneously, compare the readings to determine twist, then use math to calculate torque from twist. You could also sample the tractor side encoder for the speed, multiply that by the torque, and get shaft power. The resolution of torque would be directly related to the resolution of the encoder, the geometry of the shaft (how much it twists for a given torque), and the distance between the encoders (as large as possible). If the distance between the encoders can be increased, then lower resolution (cheaper) encoders can be used.
I am following your plan as well, but would like to offer some dissenting opinions about the method based on my personal hands on experience. First off, feel free to ignore me and try it anyways. Second, prove me wrong, because I'd love to learn new methods like you are discussing.
I believe you are planning to use a "transmissive photomicrosensor" as they are they are commonly called. These output an analog voltage (or sometimes a PWM duty cycle) which corresponds to the amount the sensor is blocked. These work, although they have a few notable drawbacks.
First off, the sensing window is small, forcing you to limit the measuring distance to a very short range (0.25" as an example). In your application, you will want to maximize the distance between the discs as to have a more noticeable twist in the shaft (which even over a good distance like 3 feet might only be a few degrees). The short distance would greatly limit the signal to noise ratio, perhaps to the point where there is no measurable twist between the discs.
Second, the sensor is an optical sensor which is very sensitive to ambient light and will report false signals from even a small amount of stray light - this will drastically reduce your signal to noise ratio. It will also be very sensitive to contamination from dirt, dust, or water.
Finally, the voltage output from these sensors will is nonlinear and will require some mapping tables in code to calibrate the voltages to the actual twist in the shaft. You'll probably end up needing to calibrate the whole thing manually and will end up with a very poor signal to noise ratio.
If you want to go this route, you might want to consider two sensors each with a single disc (just like you have pictured above) positioned as far apart on the shaft as possible. Instead of measuring the exact angular displacement as in my encoder example above, you would be comparing the timing of the light to dark transitions on each of the sensors. In this case an active photosensor with built in amplifier and comparator woud add a lot of robustness to the design. Link below.
PM-F25-P Panasonic Industrial Automation Sales | Sensors, Transducers | DigiKey
Order today, ships today. PM-F25-P – Optical Sensor Through-Beam 0.236" (6mm) Module, Wire Leads, Slot Type from Panasonic Industrial Automation Sales. Pricing and Availability on millions of electronic components from Digi-Key Electronics.www.digikey.com
By knowing the exact speed of the shaft (read one of the discs like an encoder for speed), you can calculate the angular distance traveled by the shaft during the delay between the light to dark transition on each of the discs. It would be very difficult to align the discs mechanically so each had a light to dark transition at exactly the same time under no load, so it would require a calibration cycle with the shaft running under no-load to determine a nominal delay time between each disc. Once that is defined you can load the shaft and the difference between the measured delay and nominal delay would give you the twist in the shaft. Again once you know this, a bit of math will give you torque and power. The sampling rate of the microcontroller to the sensor input would be the limiting factor in the resolution of torque. You would also need a very accurate measurement of shaft speed, so a disc with a higher pole count would be beneficial to a point although a dozen or so poles would be sufficient rather than the million pulses of the encoder idea.
The downsides are that this could not measure the torque on a non-rotating shaft where the encoder method could, and you rely on a very fast microprocessor to sample the signal for good resolution where that is not a requirement for the encoders. Also it is still an optical system so contamination is a real risk. The sensors I linked are at least waterproof.
Assume a single disk with 6 slots, with the slotted portion of the sensor radius equal to the unslotted portion, such that if we spin this disk in the field of the photointerrupter, we get a square wave with 50% ON time and 50% OFF time. Now make an exact copy of that disk and bolt the two together samely aligned. Spin it in the field of the photointerrupter and you still get a 50% square wave. Now loosen the bolt, rotate the 2nd disk by 10 degrees, retighten the bolt, spin it again in the photointerrupter field. Now you get a square wave with 33% ON time and 66.6% OFF time.
Now for a minute pretend those two plates bolted together are mounted on the PTO shaft and the offset between them represents the torque in the shaft. I will come back to this.
Consider that the period of this wave is measured by a (ex) 512kHz clock, so the raw data for the wave period looks like this (ex)
Total counts: 9482 (in one period, 1/6th of one revolution, at 540rpm)
ON counts: 3156
OFF counts: 6326
ON percentage: 33.2841%
Let's say that on the bench we measured a 14.3 degree twist at 607lbs (42.4476 ft×lb per degree), which what the math indicates our maximum safe torque is. The math then becomes:
(For this one sixth of a revolution):
Speed = 9482 clocks / 512000 clocks/sec = 0.0185195313s × 6 = 0.1111171878 seconds /rev
1/ 0.1111171878 seconds/ rev = 8.9995078151 revs/sec × 60 = 539.970468906 RPM
Torque: 50.0000% - 33.2841% = 16.7159% × 60 degrees = 10.02954 degrees × 42.4476 ft×lb/degree = 425.729 ft×lbs
HP = (539.97 RPM × 425.729 ft×lbs)/5252 = 43.77HP
And since all this is calculated during only one sixth of one revolution, we load it into a buffer and average it with the previous 5 values, and that's our output.
Ok, now we just did all that math assuming the plates are right up against each other, which doesn't mirror real life, right? Except it does. The plates will be right up against each other, and they will still measure the twist from one end of the shaft to the other. How so? Picture the shaft left to right, mount one of the disks to the right hand side. Now on the left hand side instead of fixing the plate, slide a tube over the shaft. Slide it all the way until it almost touches the disk, maybe give it a couple thou. Tack weld the tube to the shaft at the very left end of the shaft only. Now slide the 2nd disk from left to right over the tube (this disk will need a slightly larger center hole) and right up against the right-hand disk (maybe give it a couple thou spacing too) and fasten it to the tube. Now the disks are right up against each other. The shaft can't spin inside the tube because they're welded together, but it can twist inside the tube, and when it does, the angular displacement between these two disks will change because the tube doesn't also twist, as there's no torque on it.