Help solve complex math problem... reward if found.

[LVLAaron]

I'll do my best to explain what's going on here. I'm bad at machinist math and need help. I'll put a 20-dollar reward on this if someone can help solve it. I have no idea how to solve this.

I need to make a part. I have this drawing. The cone shape is 30 degrees.




Based on the drawing below.... What is the distance between point A and point B?

 

[Boswell]

why not just cad it out and read the result directly?

 

[LVLAaron]

I don't cad.

I need to cad.

But I don't know where to start.

 

[MyLilMule]

Tinkercad. It's free. Start at the beginning. You'll be more than glad you did.

 

[whitmore]

If the point-of-contact is at 30 degrees to vertical, and you know the point-of-contact to flange distance,
A to B will be
( (0.705/2 ) * (1 + sin(30 degrees) )+ (0.814 + .001 to 0.814 -.003)
sin(30 degrees), of course, is 1/2

The place to start, is to note that the sphere is tangent to the cone at the point of contact,
so that's a right angle between the radius to the point of contact and the slant surface.

 

[Martin W]

.705/2=.3525x cos30=.05437
1"+ .814-.05437=1.7596

1.7596 plus .001-.003 so 1.760 to 1.757
Just my guess
Edit. Pretty sure this is wrong, you need to find the length of the arc and I don't know how to do this.
Martin

 

[jwmay]

I think I’d solve it by contacting the designer.

 

[rwm]

This answer was wrong so I deleted it to avoid confusion. See my updated answer below.

 

[Ischgl99]

I think my math is correct on this, but if someone finds a mistake, please point it out. You would need to do as follows:

You’re using a ball smaller than the gage dimensions call for, based on the 30 degree angle and the 0.705” tangent points, the correct ball would be 1.41”. Since your ball is 1”, you need to compensate for that. This first picture finds the new tangent distance, which is smaller since your 1” ball is smaller than the design size.



Next, you need to find the change in the horizontal dimension needed, since your tangent points are closer together, your horizontal dimension will be shorter.



Now, to find your AB dimension, you need to add the radius, plus the horizontal dimension from the ball center to the tangent point, plus the original gage dimension, minus the change in length. I get an AB dimension of 1.688.



An important point, you need to make sure your new tangent points with the smaller ball is larger than your bore. I get a tangent distance of 0.5” with the 1” ball, so if your bore is that size or larger, these formulas will not work. You can do the calculations for any size ball but replacing the diameter of the ball in the formulas and going through the same steps.

 

[jcp]

My math skills are nothing to brag about but I’m going with the dimension you want is 1.666-1.670. We’ll see when the actual mathematicians show up.