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Richard MacCutchan3-Feb-20 0:32

Richard MacCutchan

3-Feb-20 0:32

Unless the devices are all measuring to the same known point in the circle you do not have any way of relating the values to each other.

OriginalGriff3-Feb-20 0:10

OriginalGriff

3-Feb-20 0:10

I know it's Monday morning, and I'm still MucusMan: The Human Hagfish but I need a diagram to work out what the heck you are asking about there ...

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Joan M3-Feb-20 0:22

3-Feb-20 0:22

https://i.stack.imgur.com/a2AyG.png[^]

2 circumferences mounted in the same center.
3 external measuring devices mounted completely unaligned with the center.
I don't know the measuring devices position.
But I get a distance from the measuring device to the circular master part mounted.
And I have 2 master circular parts.
The red lines in the drawing are the vector lines the measuring device measures would be placed into.

modified 3-Feb-20 10:44am.

honey the codewitch3-Feb-20 0:20

honey the codewitch

3-Feb-20 0:20

I'm a little confused by what you're asking, but if I understand you:

You first need the 2d Cartesian points of each radius (x,y)

Then you can do trig on them to get the angles out of it. My trig is rusty, but that's where to start.

Once you have that, you should be able to do this: trigonometry - Finding the angle between two points - Mathematics Stack Exchange[^]

Real programmers use butterflies

Joan M3-Feb-20 0:27

3-Feb-20 0:27

Yes, if I had the xy would be easy, I don't know the position of the measuring device neither the angle it is mounted...

Frown | :(

honey the codewitch3-Feb-20 0:29

honey the codewitch

3-Feb-20 0:29

Then I'm not sure what you're asking is possible. you need some way to get 2 points out of this to get an angle. That's the minimum you need (and the 3rd point is implicit)

So your task now is to divine two points. Somehow.

Real programmers use butterflies

Joan M3-Feb-20 0:31

3-Feb-20 0:31

yes...

honey the codewitch3-Feb-20 0:34

honey the codewitch

3-Feb-20 0:34

maybe try printing out your drawing and just trying to use a pen to draw the triangles you need. Then see where the points fall. From there, see where the points are in relation to the circles and see if that can't help you. It's a shot in the dark but it's the best i got.

Real programmers use butterflies

Kornfeld Eliyahu Peter3-Feb-20 0:33

3-Feb-20 0:33

What I do not understand is what distances those 3 devices measuring...
What is the significance of endpoint of the red lines in the drawing?

"The only place where Success comes before Work is in the dictionary." Vidal Sassoon, 1928 - 2012

Joan M3-Feb-20 0:35

3-Feb-20 0:35

The measured distances are the length of the red lines when they cross the circles.

Kornfeld Eliyahu Peter3-Feb-20 1:05

3-Feb-20 1:05

Like a beam that hits the surface of the circle and you have the distance without the angle?

"The only place where Success comes before Work is in the dictionary." Vidal Sassoon, 1928 - 2012

Joan M3-Feb-20 1:09

3-Feb-20 1:09

Yes, in that case is a mechanical measuring device, but yes, the red line would represent a beam that measure the distance, but I don't know the incidence angle.

Thank you for spending time on that, I'm sure my super explanations are not making it easy to understand it.

Kornfeld Eliyahu Peter3-Feb-20 1:24

3-Feb-20 1:24

Some observations:
1. Replacing the circle (disk) with an other one will not change the distance the devices measure as long as the circles are on the same plain...
2. There is an option when one or more of the devices will not hit the circle's surface (infinite or error)...
3. There is a whole sphere for each point on the circle's surface that will return the same measurement...

"The only place where Success comes before Work is in the dictionary." Vidal Sassoon, 1928 - 2012

Joan M3-Feb-20 1:27

3-Feb-20 1:27

Yes, yes and yes, but (and I'll write this in the OP) we consider all to be in the same plane.

If it is not... I simply abandon all hope and start crying now...

Kornfeld Eliyahu Peter3-Feb-20 1:33

3-Feb-20 1:33

Even the circle is on a fixed plain and the beams are always hit the surface, you can not know the position of the device as any of the devices can be at any point of a sphere and still return the same distance...
If you would know the distance between the devices it would cut down the possible position to a circle (still a lot but less the a sphere)...

"The only place where Success comes before Work is in the dictionary." Vidal Sassoon, 1928 - 2012

Jon McKee3-Feb-20 0:43

Jon McKee

3-Feb-20 0:43

With no further information I unfortunately think this is a "software can't fix your mechanical problems" situation. I'm not a math wizard though so maybe someone else will chime in.

If there's a way to guarantee a distance between the measurement device and the center of the measured sphere, I think it may be possible. You'd have both sides of the triangle coming out of the measurement device (the "true" distance [distance to center - radius] and the measured) and you could use spherical trigonometry to determine the angle between those sides (the angle of the measurement device) given the difference between the two distances. Honestly that's beyond me though and relies on having perfect spheres. I was never good at trig Sigh | :sigh:

EDIT: Also this would in no way let you know where in the xy plane those measurement devices are.

musefan3-Feb-20 0:56

3-Feb-20 0:56

Your question is really unclear, are you trying to calculate d1?

Regardless, if you have neither the positions nor the angles of the devices then you can't get an accurate calculation. Your best option would be to use which ever device gives you the shortest distance, as that will be the most accurate - but certainly cannot guarantee a correct result.

Essentially to get the correct result you need a device pointed directly at the centre point. If they are not pointing directly at the centre point then you could probably compensate for this offset is your calculations, but you would need the angles and positions to be able to do so.

Joan M3-Feb-20 0:59

3-Feb-20 0:59

I need to calculate XY position in which the sensor finds the part.

The problem is that the sensor is at an unknown position and angle (which will be very similar to the desired one but I need a really high precision (3 micrometers)).

I only get from the sensor the distance from the sensor to the part.

musefan3-Feb-20 1:03

3-Feb-20 1:03

Joan M wrote:
I only get from the sensor the distance from the sensor to the part.

So just to be clear. Do you just get a single value (distance) from the sensor to the outer circle?

Or do you get 2 values, one for outer circle and one for inner circle?

Joan M3-Feb-20 1:11

3-Feb-20 1:11

One for each circle.

Thank you for spending time on this. And sorry for my bad explanations, it's difficult for me to describe this clearly in English...

musefan3-Feb-20 1:14

3-Feb-20 1:14

Joan M wrote:
One for each circle.

Ok, that is a good start. We definitely need both. The other vital piece is: do you know the value of d1? (i.e do you know the radius/circumference of both the inner and outer circles?)

It looks like Daniel is already suggesting what I am trying to get at too, however I just wanted to be sure you have all the information that he is trying to make use of.

Joan M3-Feb-20 1:17

3-Feb-20 1:17

Yes, both circles are the master parts and we know the radius of both.

Thanks!

musefan3-Feb-20 1:20

3-Feb-20 1:20

Ok, so my trig is rusty so I will need to take some time to try and process the information.

But from what I can see I think it should be possible to do with that data - but don't hold me to that just yet, this is not a specialist area for me by far! Big Grin | :-D

Joan M3-Feb-20 1:21

3-Feb-20 1:21

Same here... trigonometry is rusty to say the least... Sigh | :sigh:

musefan3-Feb-20 2:04