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How to determine the load on the electric motor with high accuracy?
Hello. There is an engine (let's say from a screwdriver), with the help of which it is possible to measure the voltage per unit time (1/5 second), with an accuracy of 100/1000. Or maybe there is another way to determine the load during drilling to further visualize the passage of the drill through materials of different densities?
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First, you need to measure not voltage, but current.
Secondly, now dofiga current sensors with any accuracy and speed, Google to the rescue.
Third, it's pretty pointless. During the rotation of the rotor, the windings are repeatedly switched, so the current in the circuit is constantly changing.
The nature of the change in current resembles this:
If you still collect data on the current, they will be meaningless without data on the speed of rotation, as well as data on the pressing force. So think about how to add a rotation speed sensor and a pressure sensor.
Physics grade 8, power is equal to current times voltage. You can also measure the speed of rotation - it falls under load.
The load on the electric motor is the power given off by the engine, and by measuring the current and voltage, you can find out not at all, but the power consumed. Their ratio is the very efficiency that never reaches 100% and is usually not exactly known. Hence the conclusion: by measuring the power consumption, you will not know the exact load.
To accurately measure the load (i.e. power output ), you will have to measure the torque and multiply it by revolutions:
P (kW) \u003d Mcr (Nm) x N (rpm) /
9549 and if you need to get a result in hp, then it must be replaced with 7164.
But since in reality you want to measure not so much the load as the drilling parameters, then there is no need to talk about any accuracy at all. After all, you will have to somehow take into account the material of the drill, its sharpness, its wear, the force of axial pressure on the drill (and this is only the first that came to mind, and there are probably many more).
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