Digitizing arm options

[jtedmonds]

I got the information on the precision potentiometers this morning. I had asked her for a "Budget" price to see if this was an area to continue to research. The pots with the 0.007 degree resolution are approx. $135 each - not that bad considering the other options. The A/D chips are only about $5 plus a few bucks here & there. The cheapest way to determine if this thing will work is probably to build something using cheap pots and switch them out when all looks right and seems to work to get the resolution right.

On to the issue of resolving the points. I pulled out my "Vector Mechanics for Engineers: Statics" textbook this evening. After blowing off the dust I went to the vector introduction section for a brushup. Cliff, I believe you are right. I've thought about this all day and have looked at several "Vector Education" websites. The easiest way to determine the cartesian coordinates may be by converting the Spherical coordinates (distance with 2 angles - base & joint). There are fairly simple formulas for doing this. Seems to me that you could resolve the Cartesian coordinates of the joints until you got to the pointer tip or add the vectors and convert the resultant sphereical coordinates into cartesian for the solution. I think the software to run this thing should be kept as simple as possible. That should not be a problem.

I plan on doing several sketches this weekend - again mostly concerned with the math not the construction. I love a challenge, especially when it's for building something I can't justify going out and buying. Hopefully we can put our heads together and come up with a solution that is cost effective and within the skill level of most home machinists.

Vector Site - Vector Education

[Cliff_J]

I'll have to read up on the vectors, I'm pretty rusty on geometry but thanks for the research. Its been a while since I took a statics or dynamics class and the textbook for the latter was horrible enough to mess up anyone's understanding of more than the homework problems.

I think I'll order some Alps (or other) precision pots from Mouser soon and try to keep the price down to $10-20 each. Then I'll see if I can find some websites on using the soundcard as the A/D solution. Kind of a Kleinbauer approach to build as cheap as possible and then work on the precision for the second generation.

Like I said above I've got a Microscribe we can compare to. If I can build one for 1/10th the price I'll be so stoked as then I can work on building one with a large enough envelope to handle my needs instead of leapfrogging a little guy all over the place. And I could probably build the larger one with the funds from selling the Microscribe on ebay....awesome!!

I called Novotechnik as well and they have the SP2800 which is used as a throttle sensor in CART/Indy/F1 cars with .03 degree repeatability and costs $57 which is one of their least expensive pots that we might consider. I think I'll try the Alps first for the prototype and if we can get that working....

Cliff

[duluthboat]

Hi guys, I’ve been watching from the side lines, mostly because I have nothing to add in the math or electronics area. I may be able to help with the mechanics and or modeling I have looked at the Faro arms longingly but will never have the money. I’ll go back to watching, if I can be of any help just ask.

Gary

[Cliff_J]

Glad to see you're watching Gary.

What are your thoughts on constructing the machine in terms of materials and the joints and so on? And do you have ideas of other contributors on other forums that may be of assistance in the math arena?

Cliff

P.S. I just learned that most PC sound cards are high-pass filtered at around 10Hz or higher. So now we'll need to use an AC waveform or external ADC to find the potentiometer position, but this shouldn't be too hard. I just found some source code for VB that lets me read the output of the ADC in a sound card from an oscilloscope program. One more piece in our little puzzle.

[duluthboat]

The construction would be very similar to Faro, it’s about as simple as it could be. I wish I could find a nice off the shelf joint but it will likely be a machined item. How big are these encoders you’re talking about? I’m hoping the math will not be a big obstacle; we only need 5 axis positioning not 5 axis interpolation. I’m thinking of a pointer and a trigger to record a point then down load the point cloud into Rhino or TurboCad to create a surface. For myself I don’t have a need for a transducer probe.

Gary

[jtedmonds]

I did some reading today. I kept reading the term Forward Kinematics - I don't remember that from my years at Clemson. Forward Kinematics is the term for solving for pointer coordinates of an arm - at least one application. In our case, we will have known distances and angles (4). Inverse Kinematics would be resolving the angles for a target point knowing the point and arm lengths. There appears to be a lot information on the web for this subject. I did a search for Forward Kinematics and found several interesting sites. Digging through the theory for the formulas will be the hard part.

I have been thinking about the construction. I'm assuming we will machine the parts out of aluminum. I don't have a mill, but have a family friend in that business. My mini-lathe should be large enough for the parts I can make. I'm thinking about making a prototype out of very light material to see this thing work and to iron out the bugs. I was thinking that I will blow the dust off my old OOPIC for my test bed A/D converter.

Gary: I believe the P6500 (0.007 degree resolution) pots are approx. 2" in diameter.

I have recently become a HUGE fan of forums such as this. I have been very impressed with how members are very willing to help folks find the solution that best fits their needs. It's interesting to bring folks with similar interests together and pool resources, talent and experience. I think this is the best way to approach challenging projects such as this. I personally just know enough about various things to be dangerous. I also have more time to read & plan than I do actual shop time.

Since I'm new to this forum, I'm assuming a "veteran" will suggest if and when we need to move this to the Open-Source or Projects area.

[duluthboat]

I’ll do a quick and dirty model so we have something to throw rocks at. I’ll use this pot as our encoder.

Gary

[jtedmonds]

Great. I'm looking forward to seeing what you come up with. I'll be working on the math this weekend and digging out the information I have on the OOPIC. However, as Cliff said, we need all the help we can get on the MATH!

[duluthboat]

First draft.

[jtedmonds]

WOW your fast! It sure does help to see it on the screen. A rotating base would give all the degrees of freedom needed. I hope to have something together tomorrow evening on the math/formulas needed - at least some good links.

Off to bed - 5 AM will be hear before I know it

[duluthboat]

With roating base will we only need 4 axis of movement? That looks good to me.

[ger21]

Here's a basic way to find the location. There is probably a better way, but this is pretty simple. The distance between the pivots (pots or encoders) will need to be very precise. Assume that the base has not rotated and the angle is 0°. You know the position of the first encoder. Treat each segment of the arm as the hypotenuse of a 90° triangle. Use pythagorean theorem, a^2 + b^2 = c^2. You have the length(c) of the hypotenuse (length of arm segment). The base(a) of the right triangle is cos(angle of first segment) x c. With a and c, b(height of the triangle) is c^2 - a^2. You now have the position of the next encoder. Use this as the base for another right triangle, to find the next encoder position. Repeat until you get to the point. Now you'll have a height and offset from the base, just use the base rotation to get your actual X,Y,Z value. Does this make sense.

While looking for the formula to rotate the base, I just came up with an easier way. Using polar coordinates, look at the first picture and formulas here.
http://mathworld.wolfram.com/PolarCoordinates.html

Once you figure out the point(tip) location's x,y, use spherical polar coordinates to get the actual x,y,z using the base angle and the new vector length from the base to the point. See here: http://www.astro.cf.ac.uk/undergrad/...tp4/node8.html

Simple, eh? Or am I way off base here?