View Full Version : Need some math help

Uncle O
12-27-2012, 07:38 PM
I need to find 2 points on this sketch and can't seem to get past a major brain fart....I have been staring at it and .....

Anybody have a cad program and care to help me out ?
Need the 2 points shown as +, at the top where the 30 deg line would intersect to the zero line if it ran all the way up and to the right where it hits the flat on the side.. 0, 0, is the center of the sketch.
It is a .778 dia. .66 across all flats


12-27-2012, 08:00 PM
Can't happen except in one case. That is if the vertical flat is of zero length and the angle is 90 degrees.

From the X at the center to the lower right + sign is the hypotenuse of a right triangle.

The formula c squared = a squared plus bsquared. C is the hypotenuse, a is the distance from X to the right flat. b is the vertical height of the vertical flat.

If the width really is .66 and the angle really is 60 degrees, then the length of the vertial flat will be .33 and the vertical heigth will be .165 from X to the lower right +.

However, the distance from X to the 60 degree angle flat will be .381 instead of .33. The height to the upper + would be .762.

Since c squared equals a squared, then b must equal zero.

The Artful Bodger
12-27-2012, 08:01 PM
Consider a triangle from the center to the mid point of the flat and out to the 30 degree corner. That would be a right angle triangle (90,60,30 degrees) with a height of .33.


Tony Ennis
12-27-2012, 08:05 PM
I get (.33, .136) and (0, .707)

That assumes the shape is an octagon, and 60 degrees was intended.

12-27-2012, 08:15 PM
What's shown in the sketch is an octagon, but 60 degree sides would make a hexagon. With a hexagon the x on the right side would be on the horizontal centerline.

12-27-2012, 08:35 PM
I looked at this again and realized the angle from X to lower right + could be 15 degrees from horizontal.

The height of lower right + would be .088"

The height of the upper + would be .660

12-27-2012, 08:47 PM
Hi Uncle O,

*not sure* what you're aiming at! I'm guessing it's an 8-sided part, and the 30* angle is *not* tangent to the next flat around?

If the + on the right is on the corner of the flats a bit of jommetry should find that, like so:

the angle between the axis and the corner is half of 360/8, so 22.5*

so... the X coordinate is half the A/F distance 0.330"

Tan of 22.5* is 0.4142 = opposite (the Y coordinate) / adjacent (half the A/F) = Y/0.33

multiply both sides by 0,33, Y coordinate = 0.4142 x 0.330 = 0.137 (to 3 sig' fig's)

So the right-hand + is at 0.330, 0.137

To get the top + position, draw a right-angled triangle with a 30* angle at the top, the opposite side from the edge of the flats we just worked out horizontally to meet the vertical line - this will be 0.330" long (half the A/F dimension). The hypotenuse goes from our new + point to the edge of the flat and the adjacent from the intersection of the vertical line through the axis to the 30* angle.

tan 30* is 0.5773 = opposite over adjacent, so...

0.5773 = 0.330 (the half-A/F dimension) / adjacent side

multiply both sides by (adjacent) gives 0.5773 x adjacent = 0.330 (opposite)

hmmm, not there yet!

divide both sides by tan 30* (0.5773) gives adjacent = 0.330 (opposite) / 0.5773

We we now know the adjacent side is 0.330 / 0.5773 = 0.5716"

to get the position of the upper +, we have to add the Y-coordinate of the first (right-hand) +, so:

x = 0 (as it's on the vertical from the axis), Y = 0.5716 + 0.137 = 0.7086"

If I can attach it, diagram illustrates....


Hope this helps, rather than confuses!

Dave H. (the other one)

P.S. - just noticed for the first time that tan 22.5 is root(2) - 1... I'll remember that :)

12-27-2012, 09:05 PM
Your diagram is indeterminate. Here's a labeled version of your drawing that may help you sort it out though:

Uncle O
12-27-2012, 11:57 PM
Your diagram is indeterminate. Here's a labeled version of your drawing that may help you sort it out though:

Thanks for everybodies attempts to help, I have made these before, it's been a quite awhile tho.....
I need to find d as shown above, and from there I will be good. In the a.m. I will put a protractor to the corner of that flat ( on the print ) and take it from there. I will post results later.
Thanks again.

Tony Ennis
12-28-2012, 12:57 AM
I get d = .136. See my post above.

12-28-2012, 03:14 AM
I stick by my calc that d=.088.

To check it out I laid it out in Solidworks and got the same answer.

I hereby declare victory and will now run like hell.

Jaakko Fagerlund
12-28-2012, 03:43 AM
Edit: Scratch that

Paul Alciatore
12-28-2012, 03:46 AM
I have to agree with Machine. Your drawing is too rough to really know what you are asking. It could be a regular octagon, meaning 8 equal sides and 8 equal angles or it could be just a plain octagon, meaning it has 8 sides but they are not necessairly equal and the angles may not be equal. Then again, the top and bottom seem to be highly curved so it could be a figure with six flat sides and two curved ones. Or perhaps something else.

And is your 60 degree line congruent with one of the sides or is it at a different angle and only touches it at the corner marked with a cross.

In any case, you do not give enough details/dimensions to figure the location of the points. I would suggest a better sketch and more dimensions.

Jaakko Fagerlund
12-28-2012, 03:55 AM
Edited totally

The piece is round with six flats on it. Dimensions as shown, the points are here:

http://i4.aijaa.com/t/00167/11495638.t.jpg (http://aijaa.com/NWhgnj)

Highest point is 0.66, the rightmost point is 0.088 above Y axis

12-28-2012, 08:30 AM
The CAD solution is good, but here's a trig solution:-


AB = 0.33
AC = AB*cos(30) = 0.33*0.8660254 = 0.2857884
BC = AB*sin(30) = 0.33*0.5 = 0.165

ED = DB/tan(30) = 0.495

EA = ED+DA = 0.495 + 0.165 = 0.66

FG = 0.33
EF = FG/tan(30) = 0.5715768

GH = EA-EF = 0.66 - 0.5715768 = 0.0884232



Uncle O
12-28-2012, 10:37 PM
Well, I found a sample piece in my box that I had made last time around. The small flat on the side was pretty close to .180 long, so I figured the .088 dimension was going to be good. I know that sketch looked really bad, I had not intended on publishing it , just using it to work with after I had gotten home. Also when I put a protractor to that point, it reads 20 degrees, which I think I recall going thru last time I worked on these. My trig knowledge is poor, I must admit. Usually I can trod my way thru with the help of my little trig table booklet.....Anyway, the .088 dimension worked nicely, as you know. My job was to just do the flats.

Thanks again .


12-28-2012, 10:54 PM
That looks like a diamond pin for locating a part on a fixture. It probably gets heat treated, then ground on centers to a .625 shank and .750 top if it's going to be an inch size.

Uncle O
12-28-2012, 11:11 PM
Yes, it is a locator pin, no heat treat, as it is ampco. Grind as shown.

A test piece .


Jaakko Fagerlund
12-29-2012, 05:27 AM
Got me curious: How is that used to locate a part?

12-29-2012, 09:40 AM
Got me curious: How is that used to locate a part?

Part with a round hole used as a locating feature drops over the diamond pin more easily than a round pin. Common jig and fixture design technique.

12-29-2012, 10:57 AM
You have to have 2 holes in the part. You use a round pin to locate the datum hole and the diamond pin turned 90 degrees to the centerline of the 2 holes. The round pin locates the part lengthwise and both pins locate the part side to side. It is much easier to get parts on and off this arrangement than with 2 round pins and the parts are still located accurately.

Jaakko Fagerlund
12-29-2012, 01:16 PM
Part with a round hole used as a locating feature drops over the diamond pin more easily than a round pin. Common jig and fixture design technique.
Hmm, clever :) Haven't run in to them as all the jigs I've made were for die cast aluminum parts which don't have precise round holes in them.