PDA

View Full Version : Another bolt circle question...



lakeside53
01-09-2009, 12:29 AM
My circle diameter is 6 inches.
4 holes.
The distance between 2 sets of holes is 4.75 inches.
What's the other distance (2 equal) ?

If my description doesn't make sense, picture a Bridgeport J head mount..


Yes, I could draw it on graph paper, but how do you calculate the cord and coordinates when there is asymmetry? I have the boltcirc calcualor, but it's for regular spacing. Am I mssing something?

dp
01-09-2009, 01:01 AM
I guess from your description the holes form a square pattern, and that each hole is 3" from center and equidistant from each other, 90 degrees apart.

Since the circle diameter is 6" then that is the diagonal distance between hole centers. The distance between adjacent hole centers is the square root of the quotient of 6" squared divided by 2, or 4.2426", or using trig, cos(45) * 6 = 4.2426"

Don't know how you got to 4.75" so maybe there's more involved.

lakeside53
01-09-2009, 01:07 AM
No.. sorry.. my poor description. I have a rectangular bolt pattern. I know the circle diameter (6 inches), and one set of cord measurements (4.75). I need the other set of cord measurements.


What I'm looking for is a general solution to retangular bolt patterns. Then can punch coordinates into my DRO.

oldtiffie
01-09-2009, 01:08 AM
Try this answer.

Dimension required = 3.6658 or 'rounded" to 3.666 0r 3.670 (suit yourself).

http://i200.photobucket.com/albums/aa294/oldtiffie/Shop_math/Shop_math_PCD_spacing1.jpg

Paul Alciatore
01-09-2009, 01:21 AM
Quick answer: 3.6657".

I did in with CAD. Drew a 6" diameter circle with crossed lines on the center. Then offset from one of them by 4.75"/2 or 2.375" to each side. Finally measured from the intersections of the bolt circle with these two lines.

If you want to do it with math, you would envision the above construction and draw a radius from the center of the circle to one of the intersections. This would form a right triangle with the horizontal center line and the vertical line at 2.375" to one side (right for instance). The radius would form the hypotenuse of the triangle and it is half the diameter or 3". The 2.375 " dimension is the base or bottom side of the triangle and you only need to find the right side. Since a^2 + b^2 = c^2 for any right triangle and c = 3" (the hypotenuse) and the bottom side b = 2.375", you can solve for a.

a^2 = c^2 - b^2

a^2 = (3*3) - (2.375*2.375)

a^2 = 9 - 5.6406

a = 1.8329

But a is only half the answer so it must be multiplied by two.

2*a = 3.6657"

Just as found above with CAD.

dp
01-09-2009, 01:31 AM
No.. sorry.. my poor description. I have a rectangular bolt pattern. I know the circle diameter (6 inches), and one set of cord measurements (4.75). I need the other set of cord measurements.


What I'm looking for is a general solution to retangular bolt patterns. Then can punch coordinates into my DRO.

Ah - ok, Tiffie's got it for you.

lakeside53
01-09-2009, 01:31 AM
Thanks! A picture is worth a 1000 words... and... showed me where I was going to wrong (I just went downstairs to check your calculation against the item..). The Bridgeport J-head bolt hole layout isn't reactangualr.. 2 cords are the same; 2 differ:eek: A quadrangle I guess...

Hmmm... I think I'll have to resort to "measuring'. Where are my transfer points?


And Thanks Paul - I keep forgeting that I have a handy cad package.. I can use that for checking.