Math (trig) question

Matthew

Matthew

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Matthew
Started puzzling over this today. I won't be able to get back to it until sometime next week, though.

Given: A turntable, a center pivot point, and an extension shaft off to one side. At the end of the shaft is a link that is a fixed length. As the shaft extends or retracts, the link that's attached to the end of the shaft will push or pull on the turntable and rotate it within physical limits.

I don't have a good drawing package handy, so this is as good as it gets.

upload_2017-6-21_15-7-20.png

A-B1 represents the extension shaft at a "zero" position. A-B2 represents the shaft as it is extended. B1- C1 represents the link at the "zero" position, and B2-C2 presents the link at the the extended position where Bx and Cx are the attach points. D is the pivot point of of the turntable. And D-C1, as well as D-C2, is the radius.

The shaft can also retract, but I didn't show that part. When the shaft retracts, the link will pivot in the opposite direction.

I know the distance A-D. I know the radius (D-C1 and D-C2). I know the length of the link.

The angle between A-D and C1-D should be acos( A-D / C1-D). And since I already know 2 sides of the right triangle I can figure the third, subtract out the fixed length of the link, and come up with the shaft length at this particular position called "zero".

As the shaft extends, it pushes one end of the link, the link pushes where it is attached to the turntable, and the turntable rotates. The points C1 and C2 will be following the arc that is defined by the radius that pivots at point D. The B1 and B2 points will always follow the end of the shaft that can only move along a fixed line. The link is a fixed length, but its angles will change.

What I am trying to calculate:

I need to come up with a way to figure how many degrees of rotation I get as the length of the shaft increases (or decreases). For example: I need to add 5 degrees to that angle, how far does the shaft need to extend? Or, I need to subtract 5 degrees from that angle, how far does the shaft need to retract? (Angle = the angle between A-D-Cx)

I have some ideas on how to approach this, but right now they are in my head and I haven't been able to get them organized enough to put them on paper. And, as I said, it's going to be sometime next week before I can get back to it.

Any thoughts?
 
OK - forget the term "turntable" and substitute "lever". There, now it's easy.
 
Ever close a traverse?
 
Unfortunately, it's not a linear function. The change in angle will be greater at the "zero" end of the link than at the extended length. If the angle at C2 goes beyond 90 degrees, the turntable/lever will rotate back to the left. (Assuming I understand your explanation.)
 
I wrote a spreadsheet once upon a time to solve simple surveying problems. It is most certainly lost in the bytes of time.

Call the angle C2-D-A alpha. You know alpha and the side C2-D so the right triangle formed by C2-D and a segment of A-D can be fully described. The solution lies there for calculating the right triangle which can be drawn at the top of the problem.
 
upstateny said:
Unfortunately, it's not a linear function. The change in angle will be greater at the "zero" end of the link than at the extended length. If the angle at C2 goes beyond 90 degrees, the turntable/lever will rotate back to the left. (Assuming I understand your explanation.)
Click to expand...
Yeah, that will happen. As the shaft retracts, the link will swing outward because it is fixed and the radius is fixed.

It won't be linear. There should be enough known values to work with, but right now I'm looking at a 4-day weekend and my brain left a couple hours ago.
 
Clark1961 said:
I wrote a spreadsheet once upon a time to solve simple surveying problems. It is most certainly lost in the bytes of time.

Call the angle C2-D-A alpha. You know alpha and the side C2-D so the right triangle formed by C2-D and a segment of A-D can be fully described. The solution lies there for calculating the right triangle which can be drawn at the top of the problem.
Click to expand...
That's the direction I was heading. Thanks for the reinforcement. The other known angle is C1-D-A, if you want to call that beta. Beta will be known at the "zero" position, and alpha can be calculated. That leaves the angles and distances that will be up at the A-B1-C1 and A-B2-C2 corner.
 
Matthew said:
That's the direction I was heading. Thanks for the reinforcement. The other known angle is C1-D-A, if you want to call that beta. Beta will be known at the "zero" position, and alpha can be calculated. That leaves the angles and distances that will be up at the A-B1-C1 and A-B2-C2 corner.
Click to expand...
That's pretty much it. Look for the right triangles which exist and the distances (or triangle sides) which can be calculated from there. As long as it is 2D it's straightforward if a bit tedious.
 
I get the sneakin' suspicion that this draws a sine wave.
 
Matthew said:
Started puzzling over this today. I won't be able to get back to it until sometime next week, though.

Given: A turntable, a center pivot point, and an extension shaft off to one side. At the end of the shaft is a link that is a fixed length. As the shaft extends or retracts, the link that's attached to the end of the shaft will push or pull on the turntable and rotate it within physical limits.

I don't have a good drawing package handy, so this is as good as it gets.

View attachment 54401

A-B1 represents the extension shaft at a "zero" position. A-B2 represents the shaft as it is extended. B1- C1 represents the link at the "zero" position, and B2-C2 presents the link at the the extended position where Bx and Cx are the attach points. D is the pivot point of of the turntable. And D-C1, as well as D-C2, is the radius.

The shaft can also retract, but I didn't show that part. When the shaft retracts, the link will pivot in the opposite direction.

I know the distance A-D. I know the radius (D-C1 and D-C2). I know the length of the link.

The angle between A-D and C1-D should be acos( A-D / C1-D). And since I already know 2 sides of the right triangle I can figure the third, subtract out the fixed length of the link, and come up with the shaft length at this particular position called "zero".

As the shaft extends, it pushes one end of the link, the link pushes where it is attached to the turntable, and the turntable rotates. The points C1 and C2 will be following the arc that is defined by the radius that pivots at point D. The B1 and B2 points will always follow the end of the shaft that can only move along a fixed line. The link is a fixed length, but its angles will change.

What I am trying to calculate:

I need to come up with a way to figure how many degrees of rotation I get as the length of the shaft increases (or decreases). For example: I need to add 5 degrees to that angle, how far does the shaft need to extend? Or, I need to subtract 5 degrees from that angle, how far does the shaft need to retract? (Angle = the angle between A-D-Cx)

I have some ideas on how to approach this, but right now they are in my head and I haven't been able to get them organized enough to put them on paper. And, as I said, it's going to be sometime next week before I can get back to it.

Any thoughts?
Click to expand...
Are you trying to solve this problem just for fun? Does it have an application?
 
The solution requires basic algebra (solving simultaneous equations) as well as trigonometry.
 
There was a time when I had the patience for this.
 
10be89361920a9b5229200ca9aab04a3.jpg
 
spiderweb said:
Are you trying to solve this problem just for fun? Does it have an application?
Click to expand...

Real world. I use an external measurent system and calculate the angle of a part. Then I have to rotate that turntable the matching angle to make another piece of equipment parallel. I have some tricks and algorithms to do this, but this is the first time the equipment has that link, and it adds a complication to the math.
 
Matthew said:
Real world. I use an external measurent system and calculate the angle of a part. Then I have to rotate that turntable the matching angle to make another piece of equipment parallel. I have some tricks and algorithms to do this, but this is the first time the equipment has that link, and it adds a complication to the math.
Click to expand...
I can set up a spreadsheet or simple compiled program solution but it will require some consulting time.
 
Maybe it is a transcendental equation. Or maybe it has imaginary solutions (you know, i, sqrt -1). The one I like is NEGATIVE frequencies. Lots of strange math out there...
 
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"What I am trying to calculate:"

How much wood could a Woodchuck chuck................................
 
The answer is French toast. Golden fluffy French toast with butter dripping off the edges and blueberry syrup.
 
The picture is not working for me. I am drawing a blank. Can't help ya. Clueless. Trig? Whaddat? Abbreviation for trigger? LOL
 
flhrci said:
The picture is not working for me. I am drawing a blank. Can't help ya. Clueless. Trig? Whaddat? Abbreviation for trigger? LOL
Click to expand...
I gave ya all the help I could by posting pi.
 
Matthew said:
Started puzzling over this today. I won't be able to get back to it until sometime next week, though.

Given: A turntable, a center pivot point, and an extension shaft off to one side. At the end of the shaft is a link that is a fixed length. As the shaft extends or retracts, the link that's attached to the end of the shaft will push or pull on the turntable and rotate it within physical limits.

I don't have a good drawing package handy, so this is as good as it gets.

View attachment 54401

A-B1 represents the extension shaft at a "zero" position. A-B2 represents the shaft as it is extended. B1- C1 represents the link at the "zero" position, and B2-C2 presents the link at the the extended position where Bx and Cx are the attach points. D is the pivot point of of the turntable. And D-C1, as well as D-C2, is the radius.

The shaft can also retract, but I didn't show that part. When the shaft retracts, the link will pivot in the opposite direction.

I know the distance A-D. I know the radius (D-C1 and D-C2). I know the length of the link.

The angle between A-D and C1-D should be acos( A-D / C1-D). And since I already know 2 sides of the right triangle I can figure the third, subtract out the fixed length of the link, and come up with the shaft length at this particular position called "zero".

As the shaft extends, it pushes one end of the link, the link pushes where it is attached to the turntable, and the turntable rotates. The points C1 and C2 will be following the arc that is defined by the radius that pivots at point D. The B1 and B2 points will always follow the end of the shaft that can only move along a fixed line. The link is a fixed length, but its angles will change.

What I am trying to calculate:

I need to come up with a way to figure how many degrees of rotation I get as the length of the shaft increases (or decreases). For example: I need to add 5 degrees to that angle, how far does the shaft need to extend? Or, I need to subtract 5 degrees from that angle, how far does the shaft need to retract? (Angle = the angle between A-D-Cx)

I have some ideas on how to approach this, but right now they are in my head and I haven't been able to get them organized enough to put them on paper. And, as I said, it's going to be sometime next week before I can get back to it.

Any thoughts?
Click to expand...

What I'm getting from your description is that you want rotation α as a function of z, something like this. Correct?

diagram.jpg
 
I hereby request any future threads requiring super duper higher intelligence to come with a warning so that us lower people may avoid them. :)
 
This is what I came up with. I make absolutely no warranty that it is correct. I'm sure there is a more elegant solution. Been >20 yrs since I worked problems like this.
 

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register@teamandras.com said:
What I'm getting from your description is that you want rotation α as a function of z, something like this. Correct?

diagram.jpg
Click to expand...
Exactly.

And the z axis can expand or contract (I would call that figure on the left "zero position". ) I can't remember the physical limits, but probably no more than +/-30 deg and more likely half that or even 1/3 that. In real operations, much more than +/-3 deg indicates a failure of an upstream process and would rarely be seen.
 
You can find a closed-form solution for any position of the linkage by treating it as the intersection of two circles - using @register@teamandras.com's drawing, one of radius r and origin at D, and the other of radius L and origin at...well, wherever the bottom of the short linkage is in relation to D (not marked). When you boil down the math you get a quadratic equations with two possible (but not necessarily real or unique, in the arbitrary sense ;)) solutions for each . If you set the turntable center at 0,0 and an arbitrary location for the base of L it make the math a little easier. When you want to find out how much it rotates, just increase the vertical value of the origin of L and do it again. 'Cause I got that kinda time o_O

Nauga,
and 15 minutes in Matlab.
 
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nauga said:
When you boil down the math you get a quadratic equations ...
Click to expand...

I spent about 45 minutes on a white board with my college aged son who has advanced math skills that are a bit "fresher" than mine and, I think, we got to a solution. Pretty ugly looking equation...
 
register@teamandras.com said:
I spent about 45 minutes on a white board with my college aged son who has advanced math skills that are a bit "fresher" than mine and, I think, we got to a solution. Pretty ugly looking equation...
Click to expand...
I will admit I cheated first and just plotted a couple of circles to make sure it was going to work, THEN broke out the paper and derived it. If you generalize it a lot (arbitrary radii and origins for both circles) you get some ugly looking stuff but in the specific a lot is known and the mess gets down to a "simple" three-term quadratic in one axis and the other sort of falls out.

I like to poke at stuff like this from time to time to keep my skills (sort of) sharp.

It's Friday night.

Dear god, what have I become? :oops::Do_O

Nauga,
planning his next big joyride
 
Isn't there a google of mathematics where you can just post your problem and get an answer?
 
nauga said:
You can find a closed-form solution for any position of the linkage by treating it as the intersection of two circles - using @register@teamandras.com's drawing, one of radius r and origin at D, and the other of radius L and origin at...well, wherever the bottom of the short linkage is in relation to D (not marked). When you boil down the math you get a quadratic equations with two possible (but not necessarily real or unique, in the arbitrary sense ;)) solutions for each . If you set the turntable center at 0,0 and an arbitrary location for the base of L it make the math a little easier. When you want to find out how much it rotates, just increase the vertical value of the origin of L and do it again. 'Cause I got that kinda time o_O

Nauga,
and 15 minutes in Matlab.
Click to expand...

Either late this week or sometime next, I'll get back into it.

Until then, I found this:
http://mathworld.wolfram.com/Circle-CircleIntersection.html

This example shows the smaller circle with origin at (d, 0), and then changing d. In my case, the smaller circle would be at (d,y) and changing y. Wow, this brings back some bad memories from way, way long ago. I'll have to ruminate AND cogitate on this for a while.

This seems like wizardry.
 
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