Two grids, one rotating, share equivalent x-y coordinates with different values.


by Rlafrog
Tags: axis of rotation, coordinates, grids, rotating
Rlafrog
Rlafrog is offline
#1
Apr4-12, 03:36 PM
P: 1
I’m a woodworker, a math idiot, my trig hasn’t improved since I flunked it 40 years ago and I need help making a Christmas toy for my grand-kids. The values that follow are arbitrary, were extracted using eng graphics software and should be solid.

Problem: I have one 2D surface (that rotates) on a second 2D surface (that doesn’t rotate). It appears to me (?) that a given x-y point on the rotating surface can be expressed using the x-y coordinates of the non-rotating surface as a function of the degree of rotation –and also - where the axis-of rotation of the (rotating) surface lives (on the non-rotating surface).

Example:

I have a piece of graph paper with an x-y grid centered at (0,0) which I will call the Top Grid (T), and a point ‘p1’ drawn on the surface at T( 46.34, 41.69 ).

I have a piece of plywood with a grid stenciled on the surface w/ a center at (0, 0) which I will call the Bottom Grid (B).

I push a pin through the Top Grid at T(0,0) and pin it to the Bottom Grid’s point B(0,0) so that the Top Grid can rotate freely around the Bottom Grid’s center-point at (0,0). Stated differently, the axis-of-rotation for the Top Grid is point B(0, 0) on the Bottom Grid.

I align the x and y axes of both grids (rotation = 0) and mark p1 (of the Top Grid) onto the bottom grid to find that: p1(x, y) = T(x, y) = B(x, y) = ( 46.34, 41.69 ), …no surprises.

I rotate the Top Grid 10 degrees clockwise and mark p1 onto the Bottom Grid to discover that: p1(x, y) = T( 46.34, 41.69 ) = B( 52.87, 33.01 )

I separate the two grids so I can move the Top Grid’s axis-of-rotation to a different point on the Bottom Grid. Unfamiliar with Math terminology, any point on the Bottom Grid that serves as an axis-of rotation for the Top Grid that is not ( 0, 0), I refer to as the Top Grid’s “x-y offset”. I establish a new A-o-R for the Top Grid at B(46.45, -42.38) on the Bottom Grid.

I align the axes of both grids (no rotation) and mark p1 onto the Bottom Grid to discover that: p1(x, y) = T( 46.34, 41.69 ) = B( 92.79, -0.6886 )

I rotate the Top Grid 8.5 degrees counter-clockwise and mark p1 onto the Bottom Grid to discover that: p1(x, y) = T( 46.34, 41.69 ) = B( 86.12, 5.70 )

[I’m lousy at Math-speak, but I’ll take a shot anyway …] Can someone help me with a formula that I can put into a spreadsheet that will give me the Bottom Grids x-y equivalents for any x-y point on the Top Grid as a function of the Top Grid’s A-o-R ‘x and y offset’ [from B( 0, 0) to B( x, y)] with an n-degree of rotation (of the (Top Grid)?

2. Relevant equations
3. The attempt at a solution

I am attempting to solve this problem graphically, a technique that is not only prone to errors but extremely time consuming. Any help would be sincerely appreciated.
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
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jing2178
jing2178 is offline
#2
Apr5-12, 01:13 PM
P: 40
The attached file shows an image from Excel that should work for you.
Attached Thumbnails
woodworking.jpg  


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