Ball in Cone Issue: Solving for Angle X

[MechEgr]

Guys,

I have a problem I'm trying to solve for work (we're trying to inspect a hole, and I'm trying to determine its angle using a gage pin and a ball bearing).

An image of the problem is attached. The known dimensions which we can measure are shown in black, and I am trying to solve for the angle, X. I can get a value for X by plugging this into my CAD software, but I need to know how to solve for X using hand calculations so that the machinists can do it during production. The value for X should come out to be approximately 22 degrees. The four known dimensions, as shown, are D, H, T, and R. The part is symmetrical, and the sides of the partial cone are tangent to the circle.

 

[bahamagreen]

I'll give it a try...

Radius of the ball r=2.375
Cone height h= from the apex of angle X to the top of the ball.

tan(x) = R/(h-(H+T))
tan(x) = 1.5/(h-5)

and

tan(x) = r/(h-r)
tan(x) = 2.375/(h-2.375)

So, two equations in one variable, h

h=9.5

1.5/(9.5-5)=1.5/4.5=0.333...

2.375/(9.5-2.375)=2.375/7.125=0.333...

tan(18.5) is about 0.333

 

[MechEgr]

Thanks for the reply!

 

[mfb]

tan(x) = r/(h-r)

Why? The point where circle and the tilted surface meet (what your denominator should be) is closer to the apex than the center of the ball (your denominator here).

You need the sine:

sin(x) = r/(h-r)
Solve for h: $h=r+\frac{r}{\sin(x)}$
Plug this into tan(x) = R/(h-(H+T)):

$$\left(r+\frac{r}{\sin(x)}-(H+T)\right) tan(x) = R$$
Simplify:
$$r+(r-H-T)\sin(x)= R\cos(x)$$

Substitute $\cos(x)=\sqrt{1-\sin^2(x)}$, square and you get a quadratic equation in sin(x) which you can solve with the usual formula.

 

[bahamagreen]

Yeah, I was using the horizontal r but it is the angled r (at right angle to the cone slant) that is the correct one.

 

[SSWheels]

I found this method simpler (no quadratic):

Label the point at the center of the circle point O
Label the center of the bottom of the hole point D
Label the "corner" of the hole point E
Label the point where the circle touches the cone point F
Label the point a the (imaginary) apex of the cone point X, such that ∠OXF is ∠X

Connecting points O, D, E, & F gives us quadrilateral, ODEF
We know that the circle (the ball) is 0.25 from the bottom of the hole, so OD = 2.375 + 0.25 = 2.625
We know that DE = 1.5 (given)
We know that FO = 2.375 (given)
We know that ∠ODE = 90°
We know that ∠EFO = 90°

Draw line segment OE
Use trig to find ∠DEO: arctan(OD/DE) ≈ 60.255°
Use Pythagorean theorem to find length of OE ≈ 3.023
Use trig to find ∠OEF: arcsin(OF/OE) ≈ 51.772°
Angle DEF = ∠DEO + ∠OEF ≈ 112.027°

We can now find ∠DEX: 180° - ∠DEF ≈ 67.973°
This gives us ∠X: 90° - ∠DEX ≈ 22.027°

 

[SSWheels]

So, to simplify...

Given the values D, R, H, and T, calculate the following 2 values:

A=H+T-\frac{D}{2}

and

B=\sqrt{A^2 + R^2}


Then,

\angle x=\arctan{\frac{A}{R}}+\arcsin{\frac{D}{2B}}-90

or (if you prefer),


\angle x=\tan ^{-1}{\big(\frac{A}{R}\big)}+\sin ^{-1}{\big(\frac{D}{2B}\big)}-90