Simple math problem

[cfranks (Charles)]

Hi,
Could a kind person help me with what assume to be a simple math problem but it is beyond my capabilities. I want to cut a triangular goove (point first) in a metal plate, with an equilateral triangular tool, so I can imbed a 2mm cylinder with the cylinder's centre line flush with the plate. How deep must the groove be?
Many thanks,
Charles

[Wavytone]

Radius 1 mm so...

1mm x 2 x sin (60 degrees) = 1.732 mm assuming to tool is equilateral and the tool tip is perfectly "pointy". If not perfectly pointed (ie blunt) then the depth required will be a tad less.

[AndrewJ]

I get 2.00mm

Andrew

[OzEclipse (Joe Cali)]

Andrew J is correct.

The depth of the groove must be equal to the diameter of the circle.

Wavy tone has calculated the length of the sloping side of the V groove which can also be expressed as the square root of 3 x circle diameter using the old 1, 2, root 3 ratio for an equilateral triangle.

cheers

Joe

[AndrewJ]

Just to clarify Joes comment
The depth only equals the circle dia for a 60deg groove.
Basically, if the cylinder is drawn as described, the centre of the cylinder to the bottom of the groove is also the depth reqd.
If you draw a line from the centre of the cylinder to where the cylinder contacts one side of the groove, you get a line that is at 90deg to the side of the groove.
This makes a right angle triangle from centre of cylinder, base of groove and contact point.
In this scenario, depth is the hypotenuse, cylinder radius is the opposite ( ie 1mm ) and the angle is 1/2 the groove angle ( ie 30deg )
sooo by std definitions, depth * sin(30) = 1
and as sin(30) = 0.5, depth = 2
Ff you use any other angle, you need to adjust the sin(X) part.

Andrew

[cfranks (Charles)]

Thanks very much. I couldn't see that the cylinder radius met the side of the triangle at a right angle. They hadn't invented trigonometry when I was at school!
Charles

[Wavytone]

Thanks Andrew you're right. I should have drawn it before replying.