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pntrbl
02-05-2008, 11:32 PM
I need to duplicate a 3 hole bolt pattern. The linear measurement between any two of the holes, (is that called a chord?), is 3.190. How can I calculate the dia of the circle? It's been a long time since I was in school but I'm not sure I learned how to do this even then .....

SP

rantbot
02-05-2008, 11:48 PM
Law of Cosines.
http://en.wikipedia.org/wiki/Law_of_cosines

c**2=a**2+b**2-2abcosG (where c**2 = c squared)
In your case, you have c=3.190, and G=120 degrees. Also, a=b.

Carld
02-05-2008, 11:55 PM
It would be an equallaterial triangle. Your making me get out of my nice warm chair and go to the shop huh.

rantbot
02-05-2008, 11:59 PM
Continued.

So, solve for a.

c**2=2a**2-2a**2cosG=2a**2(1-cos[120])=2a**2(1.5)=3a**2
or 3.190**2=3a**2, and 3.190=a times the square root of 3, or 3.190=1.732a
Since a is the radius, a=3.190/1.732=1.842 inches.

The diameter of the circle is then twice that, or 3.684 inches.

Rich Carlstedt
02-06-2008, 12:06 AM
Or you could say the inverse cosine of 30 degrees times one half of 3.190

rich

rantbot
02-06-2008, 12:12 AM
Which is, of course, the wrong answer.

Watch those transpositions.

Rich Carlstedt
02-06-2008, 12:22 AM
Fast fingers on the keyboard ....should have said 1.8417
thanks
Rich

pntrbl
02-06-2008, 12:29 AM
Thank You gentlemen. My shop math needs improvement because both methods went right over my head. I've got 30-60-90 triangles down once again but sine/co-sine hasn't sunk back in yet.

SP

Carld
02-06-2008, 12:30 AM
:eek: there has to be an easier way.

tattoomike68
02-06-2008, 12:32 AM
This thread should make 10 pages. LOL

Paul Alciatore
02-06-2008, 01:28 AM
:eek: there has to be an easier way.

Well, if you don't like math but have a CAD program, draw a line 3.19" long. Then draw two circles with a 3.19" radius and their centers at the ends of the line. The ends of the line and one of the intersections of the circles are three points on your bolt circle. Draw a circle using those three points and you have it. List it to find the radius or diameter.

Most CAD programs will have commands for each of these steps and the math will be done by the program to at least five or six decimal places.

pntrbl
02-06-2008, 01:46 AM
I dunno about 10 pages and CAD's on the list, but 3.684" was the right answer! :) And once again I can't thank you guys enough.

What I'm doing is making new wheel centers for one of my Grandson's quads. It's been my experience that when you exceed the suspension vs terrain equation it's best if the trouble starts behind you. Bigger is definitely better up front.

6" rims all around now so I started with an 8"x3.75" trailer wheel. It had a 4 on 4 bolt pattern which seemed like a little bit too much overkill on a 165lb quad even for me. So I flamed that out, trued the cut back up on the lathe, and made new centers with the stock 3 bolt pattern. Used my Spindexer on the 3.684 dia. It fits! Here's the 1st one ready for welding. No tacks yet. Just gravity and a very nice fit.

http://i115.photobucket.com/albums/n300/pntrbl/Beau/Wheel.jpg

Still need to open up the middle to clear the hubs but that's my fixture hole at the moment.

SP

kendall
02-06-2008, 02:11 AM
Well, if you don't like math but have a CAD program, draw a line 3.19" long. Then draw two circles with a 3.19" radius and their centers at the ends of the line. The ends of the line and one of the intersections of the circles are three points on your bolt circle. Draw a circle using those three points and you have it. List it to find the radius or diameter.

Most CAD programs will have commands for each of these steps and the math will be done by the program to at least five or six decimal places.

except for the cad, I do it pretty much the same way, I just grab the compass and a pad. Hasn't failed me yet. Five minutes and I have it.

I keep trying the cad deal, but it just doesn't have the same feel.

Ken

CCWKen
02-06-2008, 02:29 AM
Now that's what HSM is all about.

Oldbrock
02-06-2008, 02:43 AM
the radius is 1.84175 providing you measured from center to center on the holes. If you measured from the inside edges of the holes that's a whole new ball of wax Peter

Astronowanabe
02-06-2008, 05:29 AM
mmmm equilateral triangles.
one thing your triangle and the circle have in common is
if you had them cut out and sitting there each would only
balance at one point. The circle at its center, and the triangle
at the point where there was equal mass in every direction
which would the common point where an equilateral triangle
was divided into three equal area sub triangles

The height of a *unit* equilateral triangle exactly
half the square-root of three (around .86603)
the centroid (balance point, center of gravity) is
a one third of the way up that height.
which would also be the center of the circle through
the corners of the unit equilateral triangle.

so the radius of the circle is the distance from
where the equilateral triangle would balance
to any of the points.

or 2/3 of sqrt(3)/2 -> sqrt(3)/3 ~> .57735

3.19 * sqrt(3) / 3 or aprox 1.84175

looks like we still have a ways to go the hit 10 pages.

J.Ramsey
02-06-2008, 06:30 AM
I use the bolt hole circle function of the DRO, takes about 10 seconds to punch in the numbers and 30 seconds apiece to pop the holes.

02-06-2008, 08:04 AM
:eek: there has to be an easier way.

Duh! Well yeah. Farm it out :D That's what I'd have to do if it absolutly positively had to be right! Math was one of my favorite subjects in school, still is, but I am no good at it now nor have I ever been.

Evan
02-06-2008, 09:57 AM
In case Astronowanabe didn't make it clear, the value .57735 is all you need to know for any size of equilateral triangle.

Simply put the sum of the shortest distance from each side to any point in the interior of an equilateral triangle is always equal to the distance of each vertex to the opposite side, the altitude. If the point is the center then all three distances are the same, 1/3 of the altitude, leaving 2/3rds of the altitude as the radius of the enclosing circumcircle. This ratio is the same for all similar triangles.

You don't need trig to solve this problem from first principles. The basic axioms of plane geometry and a simple application of Pythagoras will suffice. The rule of similar triangles is what it is all about.

http://vts.bc.ca/pics3/triangle.jpg

smiller6912
02-06-2008, 10:08 AM
I have Auto-CAD running most of the time and so I get a little math-lazy.
It's easier for me to layout three lines and a circle then just dimension the part.
No thinking involved................

torker
02-06-2008, 10:14 AM
mmmm equilateral triangles.
one thing your triangle and the circle have in common is
if you had them cut out and sitting there each would only
balance at one point. The circle at its center, and the triangle
at the point where there was equal mass in every direction
which would the common point where an equilateral triangle
was divided into three equal area sub triangles

The height of a *unit* equilateral triangle exactly
half the square-root of three (around .86603)
the centroid (balance point, center of gravity) is
a one third of the way up that height.
which would also be the center of the circle through
the corners of the unit equilateral triangle.

so the radius of the circle is the distance from
where the equilateral triangle would balance
to any of the points.

or 2/3 of sqrt(3)/2 -> sqrt(3)/3 ~> .57735