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Calculating a circle segment?

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TORQUIN

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#1
Forgive me if parts of this do not make sense, as I am not sure how to even ask what I want. I wish to learn how to calculate partial circles cut into an object.
Example:
coronetstudlargeimage.jpg


If I have a round knob already, and I want to cut in shallow radii around the circumference, like above, I need to be able to calculate the width of the circle for a given depth of that circle, or once I know the width I want to cut, what diameter cutter will I need to use to get that width while the depth is less than the radius. In the pic, the radii are very shallow, and definitely not half of the diameter of the circle that radius would be part of. Yes, I realize the pictured knob is not cut, but cast in that shape. I am thinking that once I have the circumference measurement of my knob, I then would split it into sections, like above, there would be 6 notches and peaks. I want wanted all of them to be equal width, with a circumference of 3.141 (for a 2" diameter knob), I would divide by 12, which gives me radius and peak width of .2618. Now I need to calculate the width of a circle at, let's say 40% of the radius, so I can figure out what diameter end mill to cut these radii in the knob.
Is any of this making any sense?

I have another practical application for this knowledge as well. I am essentially creating a boring head. I need to cut a notch in one side so the adjustment screw can engage the head to move it. The notch will not be half of the diameter, but maybe a quarter (half the radius), as I need the diameter of the screw to remain in the top part of the head. Learning how to calculate this information will allow me to figure out how much larger I need to make the head of the screw so I can engage the lower part of the head and still have the screw in the top part of the head. See pic below:
2018-11-05 09.57.18s.jpg

Trial and error takes time and materials. I'd rather not waste either if I can learn how to calculate this information.

Thanks,
Chris
 

RJSakowski

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#2
The easy way would be to draw it out in a CAD program. The CAD math engine does all the work.

Failing that, I would draw two intersecting circles, one the major diameter or some multiple of your knob and the second representing the cutout. You can vary the cutout circle diameter until it looks right. If you wish to see what the fluting would look like, use your compass to divide the circle into six segments and draw lines from the knob circle center through each of the dividing marks. Draw a concentric circle to the knob through the first intersecting circle center and draw circles equal in diameter to the original intersecting circle from the intersection of each of the five remaining lines with the concentric circle.

Vary the diameters until it looks right to you.
Fluted Knob.JPG
 

TORQUIN

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The easy way would be to draw it out in a CAD program. The CAD math engine does all the work.

Failing that, I would draw two intersecting circles, one the major diameter or some multiple of your knob and the second representing the cutout. You can vary the cutout circle diameter until it looks right. If you wish to see what the fluting would look like, use your compass to divide the circle into six segments and draw lines from the knob circle center through each of the dividing marks. Draw a concentric circle to the knob through the first intersecting circle center and draw circles equal in diameter to the original intersecting circle from the intersection of each of the five remaining lines with the concentric circle.

Vary the diameters until it looks right to you.
I have tried to lay something out in Fusion360, but don't yet have the skill with that program to accomplish this feat. And yes, that is the prime location to lay it all out, since I can see what it looks like and see all of the measurements at the same time.

Thanks,
Chris
 

BaronJ

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#6
Hi Chris,

I'm glad it got you what you needed ! Its many many years since I did Calculus and Trig.
 

P. Waller

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#7
If you want to know the old school method of calculating a bolt circle, what you want is a simple bolt circle where the holes are not complete.
Plunge the radii with an endmill.

I drew this as a 2" OD knob with six 1/2" diameter grooves 1/8" deep.
In order to produce a 1/8" deep groove using a 1/2" round tool the tool center would describe a 2. 25" diameter circle around the part center.

Place the part in the mill and find its center, this will be the 0 position in the X and Y Axes.

A 6 hole bolt circle is 360° ÷ 6 = 60°
Assuming the the first hole is (for lack of a better description in the 12:00 position) at 0° the first calculation will give you the first X axis position.

SIN of 0° is 0
0 X the radius of 1.125 = 0, no move on the X axis.

The COS of 0° is 1
1 X the radius is 1.125, move the Y axis ,+1.125" and drill the hole

The second axis positions are at 60°
SIN of 60° is .866
.866 X 1.125" = .974", move to this position in the X axis.

COS of 60° is .500, R1.125" X .500 = .563", move to this position in the Y axis and drill hole.

Continue around the circle adding 60° to each position until finished. Be aware that positions on the other side of 0 will be negative numbers.

Simple calculator work really. This is how it was done pre-computer, in more modern times the machine control or cad program merely does the math for you from a description of the hole circle diameter and number of holes required.
Knowing how to do this is for the most part useless knowledge in a shop today.
An inexpensive DRO will do these calculations for you as well using the Bolt Circle Function.

If you have a manual mill without a DRO you will have to do it yourself.

Good Luck

 

TORQUIN

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#8
I found my answer in the Sagitta formula. I know my chord length, as I chose it based on the circumference. The sagitta formula gives me the height of the chord within my circle, knowing how long I want that chord to be.
http://www.mathopenref.com/sagitta.html
In this case I want equal lands and scallop cuts around the circumference so that it looks symmetrical. My part is 2.750" diameter. The cutter chosen for this is a 1.5" end mill.
2.750" x 3.1416 = 8.6394"
Divide into 12 equal lengths, 6 land and 6 scallops: .7199"
Using the formula I got the depth of cut (height), touching off from edge. This was the depth of cut for the cutter diameter, not the part, as the circumference involved was 1.5". This is the depth of the 1.5" cutter at which it measures .720" on the chord line.
DOC (height) = .750 - SqRt(.75^2 - .360^2)
This gave me a .092" DOC.
Tests on this proved to not give me the result I wanted as I had not accounted for the arc of the part. That made the cut end up too narrow, about .660" in stead of the desired .720".
So, I ran the same calculation for the part also, and added the DOC for the part to the DOC of the cutter. Remember, we are getting the depth for a given arc where the chord length is X (.720"). For the 2.750" part, the value is .047". Added to the cutter's .092" gives me .139" DOC.
Touching off from the edge of the 2.75" part and cutting to a depth of .139" with the 1.5" cutter gave me lands and scallops of equal length, at least as best I can measure them.
2018-11-07 17.57.11s.jpg

I retested this with a chunk of plastic which is 3.824" diameter, which comes out to 12.013" circumference. I used the same 1.5" end mill, as I thought the depth of the scallops on the 2.75" part was not enough for my liking. Using the formula on the larger part and the same cutter, I got a DOC of .247". I got 1" scallops and lands, as best I can measure.

2018-11-09 17.47.40s.jpg
2018-11-09 17.48.18s.jpg

I was measuring more accurately with calipers, but even they are hard to hold exactly in place even when not holding the camera.
The cutter choice ended up being 54% of the 2.75" part, and 40% of the 3.824" part. On the larger part the scallops are a bit deeper and, to me, they look a little better.

Chris
 

RJSakowski

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#9
As you have discovered there, are an infinite number of solutions to the conditions where the major diameter of the knob is 2" and the chords (distance across the arcs)around the the knob are equal in length to the chords of the flute. This is why I suggested drawing out the knob to decide what is most aesthetically pleasing. SolidWorks, Fusion, or other parametric CAM program make this a few minute task.

If you decide to brute force the task, it can be done that way as well. The equal chord requirement means that you will have a twelve sided polygon (dodecagon) with equal facets. If you have a rotary table, the machining is quite simple. Center the RT on the mill table and select and end mill and mill the 2" disk by rotating the table. I usually choose a displacement from the RT center in the x axis only to simplify the calculation. The y position is zero. The x position is determined by the knob and flute diameters in the derivation below.

To cut the flutes, I would select an end mill diameter which will provide a unique solution. .75" seems like a decent choice for the end mill. I would move the end mill to the offset position and plunge cut the first flute. Then I would rotate the RT 60º and cut the second flute. Rotate and cut until all six flutes are cut.

Without an RT, the flutes can still be cut. You just need to determine the coordinates of the four flutes that are off the x axis. With a DRO, again this is simple. Just set up a bolt circle with the radius equal to the offset distance. Without bolt circle capability, you can calculate the sine and cosine of the offset distance. The absolute distances in x are the cosine of the offset and in y are the sine of the offset. The four coordinates are the + and - applied to that distance.

Personally, Unless there was an absolute requirement for the exact same chord length, I would just eyeball it. That task can be done by construction with a piece of paper and a compass in a couple of minutes. Play around with the diameters until it looks good and measure them with a rule. You might want to record your numbers if you may have a need to duplicate the knob in the future.
-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
Cutting an equal chord six sided fluted knob

Some defining terms:
D = major diameter of the knob
d = the diameter of the flute
R = D/2 radius of the knob
r = d/2 = radius of the flute
C = the length of the chord starting and ending on the circumference of the knob and spanning the uncut portion of the knob
c = the length of the chord starting and ending on the circumference of the knob and spanning the flute
L = the distance from the center of the knob to the chord C
l = the distance from the center of the flute to the chord c
A = the angle subtended by the chord C
a = the angle subtended by the chord c

By the original condition of 12 equal chords, L = l and the polygon described by the chords is a dodecagon. The angle A is 360º/12 or 30º

The problem is determining the offset from the center of the knob to the cutter center. Since an equilateral dodecagon is described by the chords, chord C and c are the same distance from the center of the knob and C + c is the distance of the cutter center to the center of the knob.

The relevant equations are:

c = 2 r sin(a/2), C = 2 R sin(A/2)
l = r cos(a/2), L = R cos(A/2)

For a 6 flute, equal chorded knob, A/2 = 15º and sin(15) = .2588
For a 2" diameter knob, C = 2*.2588 = .5176"
Since the chords are equal length, c = C =.5176
and L = 1" * cos(15º) = .9659"
The offset distance is L + l = .9659 + l

The pertinent equations are, l = r cos (a/2) and c = 2r sin (a/2)
Solving the second equation for a/2, a/2 = arcsin(c/2r) = arcsin(C/d)
And l = r cos(arcsin(C/d))
In generic terms, the offset distance is L + l = D cos(A/2) + d/2 cos(arcsin(D/d *sin(A/2))

Assuming a flute diameter of .75 and plugging in some numbers,
L + l = 2 * cos(15º) + .375(cos(arcsin(2.6667*sin(15º))
or L + l = .9659 + .375 * cos (43.64º) = .9569 + .2714 = 1.2283"
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TORQUIN

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#10
Sorry I did not mention before, I am doing this with with a dividing head.
Here is the actual part I want to cut the scallops in, for holding my tree branch pruners on my tractor and lawn mower.
2018-07-14 08.11.32s.jpg

2018-07-28 10.14.44s.jpg

Thanks,
Chris
 
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