Well i'll be darned...

After posting this "quick message", I saw an "edit" link, I edited this message an... up poped an image upload slot.

It's amazing how web site builders overlook the simple things!

Here is the Image!

Only the distances marked with the text "known" are known.

The rest is placed there for reference.

EDIT: Wait, just use the MathsIsFun uploader - I think it can upload images.

Is there such a thing?

]]>Give/send an email address and I will send it to you.]]>

[img]e:\crom\arc-to-arc.jpg[/img]]]>

If you can create one, you can then upload it (click "Post reply" then in the "Image Upload" section, select 1 or more upload slots, find the file using browse and click submit)

]]>The raidus is 1329.0000000

The start angle is 68.45736452

The end angle is 90.00000000

The arc-length is 499.69060185

Also from "P1" horizontally back to the left (angle=180.000000), to a vertical 90 degree line from the end-angle of the 1st arc is 488.0000000 long.

This is how I found my starting point (P1) on the 1st arc.

At the start-angle (P1) a vertical line is 28.875000 long. (This length will never change)

Angle = 90.0000000

At the end of this line (call this "P2") is the start of another arc going counter-clockwise.

The radius is 77.15944660 (This radius will never change)

Back to the center-point of the arc is:

The Angle in XY Plane = 277.43909137 (from "P2")

Delta X = 9.99000000

Delta Y = -76.51000000

I need a calculation to find the intersection of these 2 arcs. (in terms a dummy like me can understand)

I hope someone can help!

Thanks,

Dan