I worked a little on electronics at school but it's a long time since then so I need some advices.

I want to track a finger pressure position on a string on the X axis (see the figure below). I think that I can deduce it from the difference of forces on the 2 sensors, but I'm not sure this will work.

Schema

So, is this method correct and what kind of sensor can do the job ? Or do you have any other method that'll work ?

Thanks.

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Isn't this actually a mechanical problem disguised as electronic by talking about a sensor? – Federico Russo Jul 22 '11 at 14:34
Yes it is absolutely mechanical. The question is about validity of deferences in forces. Also it is geometry question. And after all above it is arithmetical problem. Electrical engineers are bad helpers with this matters – Rocket Surgeon Jul 22 '11 at 15:39
Do you actually want to track a finger's position, or do you want to know the pitch of a plucked string, or perhaps the force applied? – Joby Taffey Jul 22 '11 at 16:18
Thanks for your answers ! :) In fact, I'm currently thinking of a game concept which involve plucking a string at different positions (not at the same time of course), and I didn't know if that was possible. So, I think, I'll give a try with the olin's method. – Mr_Qqn Jul 23 '11 at 10:11
Is the string being hit or does it need to measure while being touched continuously? – endolith Jul 26 '11 at 21:06
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6 Answers

up vote 3 down vote accepted

From your drawing, it looks like the string deflects relatively little relative to its length. It also seems there may be accidental sideways friction force on the string from the finger. Both these together mean that you do not want to measure the X component of the force at the anchors.

There are two reasons why measuring the X force won't give you useful information:

  1. It's a string, so the force along is axis is the same everywhere (assuming no finger friction for now). The X component of that force at each anchor will be the cosine of the angle at which the string leaves the anchor. We'll define zero angle to be when the string leaves horizontally, which is essentially the unloaded position. This angle is expected to be small, so the cosine will be nearly 1 for all cases. Not useful.
  2. Accidental sideways force from the finger on the string will show up predominantly in the X forces at the anchors.

What you want to measure are the Y forces at each anchor. To calibrate, measure them with no load, then subtract that off from subsequent readings. All the vertical force on the string must ultimately be ballanced by the sum of the vertical forces holding up the anchors. Sideways friction forces only effect the vertical force by the sine of the angle, which is near 0.

Strain guages or capacitive scales are probably a reasonable way to measure the vertical force. The rest is a little circuitry to convert the signal appropriately for a microcontroller A/D input, and the rest is firmware.

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Consider also (trying to) measure the deflection angles. X1/X2 = Angle2/Angle1 ie actual angle values in radians or degrees. – Russell McMahon Jul 22 '11 at 12:21
@Russell - if both angles are very small (which is not unlikely) the problem is ill-defined, and thus the error on the X1/X2 ratio may be huge. – stevenvh Jul 22 '11 at 12:33
To be more specific, it's not the small values themselves which make it ill-defined, but the lack of precision which follows from that. – stevenvh Jul 22 '11 at 12:39
@Russell: In theory, the deflection angles do tell us the X position fo the finger. The Y force component at each anchor to cancel the finger's downward force will be proportional to the sine of the angle, so I think we're mostly agreeing. – Olin Lathrop Jul 22 '11 at 12:46
Thanks for your explanations. As stevenvh said, the angles may be small, but I will I least give it a try and check what I can get from. – Mr_Qqn Jul 23 '11 at 10:17
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No, that won't work. If the pressure is vertical, as drawn, then the X-components of the force on either side will be equal. If you draw a vector diagram of the forces this will be clear. If the vertical movement of the finger is limited then both change in stress in the string, and Y-components of the force will be too small to make a proper measurement (if you can measure the latter in the first place).

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There will be a difference in force on each side, unless the string is very straight. – JakobJ Jul 22 '11 at 12:12
@JocobJ - Yes, but it will be in the Y-direction only. Y-direction may be difficult to measure to start with, but the Y-component will also be very small if the string deflection is small compared to its length. – stevenvh Jul 22 '11 at 12:17
@Jakob: No, there won't be a difference in the force magnitude on each side, at least if we ignore sideways friction forces from the finger. That's what a string does, it has the same axial force everywhere in a free span. The difference in the force is due to different direction, which will be most apparent in the Y component of the total force at each end. – Olin Lathrop Jul 22 '11 at 12:50
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Description was not too bad but, as ever, a betterdescription of what you are really trying to do will help. eg is this a guitar string or ... ?

  1. Olin's idea looks goodish.

  2. Consider also (trying to) measure the deflection angles.

X1/X2 = Angle2/Angle1

Xx = distances along X axis. Angle = deflection angle of string from horizontal. ie actual angle values in radians or degrees.

If you connect the strings to stiff pivoted rods at the connection points or eg take strings through holes in pivoted capstans, you can measure either the angles directly or if you apply rotational restoring force proportional to rotation, the rotational restoring force can be measured.

IF the string is above a surface then a detection of the finger may be better. This could be optical or capacitive or ... .

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One possibility is to measure the acoustic waves that travel down the string when it is touched. Clamp each end to a vibration sensor, like a piezo, and they will produce a pulse when the wave hits them. If the finger is exactly in the middle, the pulses will occur at the same time. If it's 1/3 of the way, one pulse will occur at time T, and the second will occur at time 2T. etc. I believe this is how some touch-screens work.

This will only detect taps or changes in where the finger is touching. For continuous touch, you can assume that the previous touch hasn't changed, since changing the position or removing the finger would (I think) generate new pulses.

Alternately, you could get fancy and do time-domain reflectometry on a conductive string. Send high-frequency pulses into the string from one end and measure the delay until you get a reflection from the finger. If you only get an echo from the other end of the string, then you know it's not being touched.

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Yes, you should be able to measure the x-position based on the force on each side of the string. I have used strain gauge sensors to measure force. They are pretty cheap and relative easy to use. I would recommend placing the two strain gauges in a H-bridge and use a instrumentation amplifier to get the best result. (noise and temperature immunity)

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The method is correct. The sensor mechanics must be designed to measure the tention force only exactly to current direction of the rope. For example with some recalculation for motion of real sensor, like fisherman weight scale, you can derive exact total length of rope including 2 deltas from 2 reading of springs.

The geometrical rules for X, Y vectors:

  1. Y1 + Y2 = Total pressure
  2. X1 -X2 = 0
  3. F1 = \$\sqrt{X1^2 + Y1^2}\$
  4. F2 = \$\sqrt{X2^2 + Y2^2}\$

etc.

The major mechanical requirement is to have sensors directionality to comply with vector. I.e. Sensor must read only tension on string, not the flex of the joint, or torque of string rotation.

The circuitry choice will be next step after you choose the sensor (completly impossible to choose before you decide what kind of sensor, cost, accuracy, noise immunity, reliability you need.

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