Cutting a simulated curved groove. Confirmation of formula for head tilt angle.

[Parlo]

I have posted about this before, referring to it as "faking a radius". The handwritten note that Fred gave me says "Dia of Radius to be cut Divided into Dia. of Cutter=Sine of Angle" followed by "Cutter must be smaller than Dia of Radius to be cut"

I understand this to mean cutter diameter divided by arc diameter.

I used this once many years ago. It was adequate for the situation.

That's the formula I've looked at but the elliptical arc was way off - see my original post for the error.

 

[MrWhoopee]

That's the formula I've looked at but the elliptical arc was way off - see my original post for the error.

It appears that you are expecting the arc to intersect a true radius at 3 points (both ends of chord and center of arc). I believe it would have to be a true radius for that to occur. When I used it, I cut to width and depth was whatever it was.

 

[Parlo]

It appears that you are expecting the arc to intersect a true radius at 3 points (both ends of chord and center of arc). I believe it would have to be a true radius for that to occur. When I used it, I cut to width and depth was whatever it was.

The idea is based on that for any 3 points where the centre point is not on the same line, an elliptical arc and a circular arc can pass through all 3 of them. I know where the points are from the circular arc, it's finding the ellipse that fits them too. I know the diameter of the cutter I just need to find the angle. I can't find a formula online that gives this result.
I'm sure my calculated angle (using the depth as a variable) of 19.47 degrees will work. I wanted to see if there was an alternative formula that probably also takes the depth into account.
I'll cut it tomorrow & see.

 

[twhite]

I have used this many moons ago.

Cutting oil is my blood.

 

[Martin W]

The two points along with the center of the circle form an isosceles triangle. Two sides are radii of the circle. The base of the triangle is a line segment connecting the two points. Bisect the base with a line segment from the base to the center of the circle. Now we have two triangles, each with hypotenuse being a radius of the circle and the base being half the distance between the two points. The angle is half the angle that we want. sine of the half angle is the opposite side of the triangle (half the distance between the two points) over the hypotenuse (radius of the circle). The solution is then

angle = 2 x arcsin (0.5 x |P1 - P2| / radius)
Martin

 

[aliva]

Here is a sketch.

You don't give any dimensions but a boring head will give you the same result.

 

[Parlo]

The two points along with the center of the circle form an isosceles triangle. Two sides are radii of the circle. The base of the triangle is a line segment connecting the two points. Bisect the base with a line segment from the base to the center of the circle. Now we have two triangles, each with hypotenuse being a radius of the circle and the base being half the distance between the two points. The angle is half the angle that we want. sine of the half angle is the opposite side of the triangle (half the distance between the two points) over the hypotenuse (radius of the circle). The solution is then

angle = 2 x arcsin (0.5 x |P1 - P2| / radius)
Martin

That's great Martin.

Can you please substitute the dimensions I gave and the resulting angle please.

 

[Parlo]

You don't give any dimensions but a boring head will give you the same result.

The radius to be approximated is 100mm @ 5mm deep with an 80mm diameter cutter.

 

[Martin W]

My figure comes up as 15.5 degrees. Words of warning, the kid in the meme was me when I was nine.Lol
Martin.

 

[twhite]

My figure comes up as 15.5 degrees. Words of warning, the kid in the meme was me when I was nine.Lol
Martin. View attachment 451509

Awesome. How large of projectile did you ?


Cutting oil is my blood.