Math question...

[jhanko]

My math just isn't what it used to be, and wasn't very good at it's best! Can someone solve this? Imagine the circle is a cross section of a light and the rectangle is a tritium vial. How deep would I have to cut into the light so the left & right upper edge of the trit is flush with the OD of the light?

 

[Alan B]

My math just isn't what it used to be, and wasn't very good at it's best! Can someone solve this? Imagine the circle is a cross section of a light and the rectangle is a tritium vial. How deep would I have to cut into the light so the left & right upper edge of the trit is flush with the OD of the light?

Make a right triangle from the center top of the installed vial to the right tip of the vial at the edge of the light to the center of the light, and solve this triangle:

If L is the length of the vial
and R is the radius of the light

R^2 = (L/2)^2 + X^2

Solve for X

X = SQRT(R^2 - (L/2)^2)

Where X is the top of the vial to the center of the light (the height of the right triangle)
and L/2 is the base of the right triangle

So cut depth is: R - X + H

Where H is the height of the vial

Just noticed. This is post 1,000 for me.

 

[precisionworks]

solve this triangle

Good call, Alan:thumbsup:

This question is nearly identical to a keyway depth question, and both can be quickly trigged out. Trig is amazingly useful in a machine shop, and there are entire chapters in Machinery's Handbook on the different formulas & their applications. If you learn how to solve for "the big six" functions, the rest is easy.

Those six? Sine, cosine, tangent, cotangent, secant, cosecant. Those formulas are on page 51 of Machinery's Handbook, 27th Edition.

Didn't mean to get so far OT

 

[jhanko]

Make a right triangle from the center top of the installed vial to the right tip of the vial at the edge of the light to the center of the light, and solve this triangle:

If L is the length of the vial
and R is the radius of the light

R^2 = (L/2)^2 + X^2

Solve for X

X = SQRT(R^2 - (L/2)^2)

Where X is the top of the vial to the center of the light (the height of the right triangle)
and L/2 is the base of the right triangle

So cut depth is: R - X + H

Where H is the height of the vial

Just noticed. This is post 1,000 for me.

I'm sorry, but I just can't figure it out...

 

[Alan B]

I'm sorry, but I just can't figure it out...

Start by taking the diameter and dividing by two to get R.

Take the length of the vial and divide in half to get L/2.

Square R to get R^2

Square L/2 to get (L/2)^2

Subtract to get the difference R^2 - (L/2)^2

Take the square root to get X

Subtract X from R to get the depth to the top of the vial

Add the height of the vial to get the total depth

That work?

 

[jhanko]

Start by taking the diameter and dividing by two to get R.

Take the length of the vial and divide in half to get L/2.

Square R to get R^2

Square L/2 to get (L/2)^2

Subtract to get the difference R^2 - (L/2)^2

Take the square root to get X

Subtract X from R to get the depth to the top of the vial

Add the height of the vial to get the total depth

That work?

I think so. I come up with .073". Is that correct?

 

[AuroraAlpha]

I think so. I come up with .073". Is that correct?

Cad says...

...pretty much.

 

[jhanko]

Cool! Thanks for the free math lesson and for not giving me the answer, forcing me to use my brain...:hairpull:

 

[mdocod]

Also....

For convenience, there are numerous triangle calculators on the net that will save time punching away at a calculator. Just make sure it's one that goes out a few significant digits, they vary.