# One for MKlotz - Calculate Cut of 90 Deg V ### Help Support HMEM: #### dwentz

##### Well-Known Member
OK I am working on a project and need to cut a 90 Degree V in a part .133 Deep.

I have the part to be machine set at a 45 Degree angle and have the corner of the mill set at the center line of the cut.

What I can not figure out is how deep and how for back do I need to move the part to get the correct depth of .133

I have done this before by trial and error, but it would be nice to get this one correct. #### rkepler

##### Well-Known Member
What's the size of the square in there? I recall 3/16 and if I remember how I did it from that position I started with the edge dead on center, fed the Z 1/2 of the 3/16 and then fed the X (or Y) the other half. That should give you 2 faces each 3/16".

Not sure where the .133 came from...

#### maverick

##### Well-Known Member

For .133 depth ,the move is Y+.094 and Z-.094,
.094 is the base and height of a right triangle whose hypoteneuse is .133

#### bezalel2000

##### Well-Known Member
IIRC
multiply .133 x Sqr root of 2 then divide by 2

or .133 x 1.4142 / 2 = .094

z= down .094
x= .094

Bez

#### mklotz

##### Well-Known Member
You've already had several correct answers assuming that the 0.133 dimension is the depth of the V from the surface to the deepest point. Then the answer is simply:

0.133 * sin(45) and 0.133 * cos(45)

Since sin(45) = cos(45) = 0.5*sqrt(2) = 0.707...

the answer is 0.094 in both cases.

It's not always easy for the responders to visualize the details of a question like this. In the future a diagram will help to ensure correct answers.

#### dwentz

##### Well-Known Member
Thanks everyone, I looked in the machinery handbook but could not find it. I was going to sneak up on a layout line, which I had done in the past, but I wanted to know the correct way as I have run into this before.

Thanks again,

Dale