Measuring rotational speed for Tachometer vs Oscilloscope

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Discussion Overview

The discussion revolves around the discrepancy in rotational speed measurements between a tachometer and an oscilloscope, specifically addressing why the oscilloscope appears to double the rotational speed indicated by the tachometer. The conversation includes technical explanations, hypotheses about the measurement setup, and potential corrections to the measurement process.

Discussion Character

  • Technical explanation, Debate/contested, Experimental/applied

Main Points Raised

  • One participant questions why the oscilloscope shows a rotational speed that is double that of the tachometer, providing specific measurements from both devices.
  • Another participant requests details about the experimental setup, including the type of tachometer and motor used, suggesting that the tachometer might be locking to a harmonic of the rotation.
  • It is noted that the hall-effect sensor may be sensing only one polarity of the magnet, which could affect the readings.
  • One participant asserts that the discrepancy arises because the sensor outputs two pulses for every rotation, implying a need to adjust the oscilloscope's gain to match the tachometer's reading.
  • A later reply challenges the assumption that the two pulses per rotation are equally spaced and suggests measuring the period of the signal to determine frequency accurately.

Areas of Agreement / Disagreement

Participants express differing views on the cause of the measurement discrepancy and the appropriate adjustments to the oscilloscope settings. There is no consensus on the exact nature of the problem or the solution.

Contextual Notes

Participants have not provided a complete description of the experimental setup, and there are unresolved questions regarding the specific characteristics of the tachometer and motor being used. The discussion includes assumptions about the behavior of the hall-effect sensor and the nature of the output signal.

Special One
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Homework Statement
f = 64.3 [Hz]
w = 1930 [RPM]
Relevant Equations
w=2*pi*f
Question: Why does the oscilloscope double almost the exact value of rotational speed measured by Tachometer?

Rotational speed from Tachometer = 1930 [RPM]
Frequency of 1 period = 64.3 [Hz] which means 3857.91 [RPM]

The output waveform of hall-effect sensor is attached.
Can you have any explanation please?
 

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We need a link to the experimental setup.

A magnet has two poles, but it appears the oscilloscope is seeing only one polarity, so that should be OK. What is the hall effect sensor sensing? How is the magnet mounted?

What type of tachometer are you using?
Maybe it is locking to a harmonic of the rotation.

What type of motor is it?
At 65 Hz, it is probably not an AC induction motor.
The distributor of a 4 stroke motor, where there might be a Hall effect sensor, rotates at half the speed of the crankshaft.
 
Baluncore said:
We need a link to the experimental setup.

A magnet has two poles, but it appears the oscilloscope is seeing only one polarity, so that should be OK. What is the hall effect sensor sensing? How is the magnet mounted?

What type of tachometer are you using?
Maybe it is locking to a harmonic of the rotation.

What type of motor is it?
At 65 Hz, it is probably not an AC induction motor.
The distributor of a 4 stroke motor, where there might be a Hall effect sensor, rotates at half the speed of the crankshaft.
The answer is because it puts out 2 pulses for every rotation.
So, in order fix this problem we should change the gain in oscilloscope from
+60.000,000 to +30.000,000
Now we will be getting the exact value as it is in the tachometer.
 
Special One said:
The answer is because it puts out 2 pulses for every rotation.
What is "it", and why does it "put out" two pulses per rotation ?

Special One said:
So, in order fix this problem we should change the gain in oscilloscope from
+60.000,000 to +30.000,000
The gain of an oscilloscope sets the vertical scale, not the sweep rate.

I would not assume that the two pulses per rev are equally spaced. So you should measure the period of two cycles of the signal with the oscilloscope, then take the reciprocal to convert that to frequency.
 

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