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

MrWhoopee said:
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.
Click to expand...
That's the formula I've looked at but the elliptical arc was way off - see my original post for the error.
 
Parlo said:
That's the formula I've looked at but the elliptical arc was way off - see my original post for the error.
Click to expand...
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.
 
MrWhoopee said:
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.
Click to expand...
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.
 
I have used this many moons ago.
bdcfd1f7193e5e9d825cee370409b1e3.jpg

3d2f642a32f7d2e884dd80e44f455b47.jpg

c501778a5cbdf68d7dbaae2b584d3672.jpg

3d3344e9674bf89a5cbf71bb7a7733de.jpg

3a4d2b85e47fa8853ed5377fa0da0d4b.jpg

d54944117fff9644f5921b5cab51ecf9.jpg



Cutting oil is my blood.
 
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
 
Martin W said:
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
Click to expand...
That's great Martin.

Can you please substitute the dimensions I gave and the resulting angle please.
 
You must log in or register to reply here.
Back
Top