1. Member
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Compound angles

Been struggling with this for a while now.

Having a rectangular block I want to mill a groove whose bottom direction is inclined to all sides at angles A, B & C (angles measured in the planes of the sides of the block).

If I am correct the first two steps are:

With the sine vice jaws vertical clamp the block rotated in the vertical plane of A such that top surface makes angle A with the vice base plane.

Tilt in the vertical plane the sine vice jaws to angle = atan (tanB x cosA)

Now comes the part I can't figure out. I need to set the vice base at some
angle in the X - Y plane (a function of C, A and B) to the X axis of the mill. What is this function?

Help, corrections much appreciated.

John.

Cross posted to RCM.

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Can you repeat that?

That stuff can get really confusing, short circuits the thinking process.

oldtiffe.......where are you, he's good at this stuff.

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Compound angles.

Originally Posted by machinist60
Been struggling with this for a while now.

Having a rectangular block I want to mill a groove whose bottom direction is inclined to all sides at angles A, B & C (angles measured in the planes of the sides of the block).

If I am correct the first two steps are:

With the sine vice jaws vertical clamp the block rotated in the vertical plane of A such that top surface makes angle A with the vice base plane.

Tilt in the vertical plane the sine vice jaws to angle = atan (tanB x cosA)

Now comes the part I can't figure out. I need to set the vice base at some
angle in the X - Y plane (a function of C, A and B) to the X axis of the mill. What is this function?

Help, corrections much appreciated.

John.

Cross posted to RCM.

Please make a sketch, scan it, post it to PhotoBucket (or similar) and post the link to this thread and I will see what I can do to help (thanks Ken - appreciated).

It will also help if you can advise if you mill head can tilt or not and what tilting vices and rotary tables you have.

Compound angles are difficult (3 planes) but are a lot easier if they can be successively reduced to one or two planes - preferably "X" and "Y" or "X" and "Z" or "Y" and "Z" etc. as they then relate directly to the mill axis. "Rotation" can be dealt with by a rotary table or a rotating base on a vise.

Here are two versions of one I did some while ago here about tool settings as examples:

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I did a drawing to make the angles a, b, and c plain. The bolder line represents the bottom of your groove and the angles are measured from the dashed lines as shown.

So, what you said about angle a is correct. If the front and back of the block in my drawing are against the vertical jaws of the vise, then the block should be tilted to angle a in the vise as measured between the bottom of the block and the bed of the vise.

Now it gets a bit more complicated. Since it is now tilted at angle a in the vise, the line that represents the bottom of the groove will appear to be slightly longer as viewed from directly above. Actually, since it is now horizontal in reality, it will appear in it's actual length in the horizontal projection as viewed from above. So, if we simply rotate the vise by angle b, the cut will be at the wrong angle. What we need to do is find the angle that the horizontal projection of the tilted line must be set to to produce the desired actual angle. Some actual trig is needed.

I need to consult my math reference book, which is at work. I will continue this tomorrow with an exact solution. If anyone else wants to jump in, please feel free to use my drawing if you wish.

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OK, it took me a little longer than I thought. But the answer is easier than I suspected.

As I said before, your first step of mounting it in the vise with a rotation by the front view angle is correct. Here is a 3D view. Only the bottom of the groove is shown for clarity. The front face would be the side against the front vise jaw.

When the part is rotated by angle a, the bottom of the groove becomes horizontal. Also note that the length in the X direction is increased slightly. This is shown both here and in the second drawing.

Here we see a front view. Since the projected length (0-A' or Lxr) is the same as the projected length before rotation (0-A) this length is given by a simple trig formulae. cos (a) = Lx / Lxr or as transposed in the figure, Lxr = Lx / cos(a).

Now we view it from above.

Again, 0-A is the original, unrotated position of the groove and 0-A' is after the rotation. The increase in projected length is shown as the short line A-A'. It has been stretched a bit for clarity. First we solve for the distance y using the original projected length, Lx and the tangent of the angle b. Then we solve for angle b' by using y and Lxr. Substituting the values for these two quantities and a bit of manipulation and we get the final equation at the bottom: b' = arc tan (tan(b) * cos(a)). I forgot the final bracket in the drawing, but the correct form is here.

This angle, b' is the amount you rotate the vise on the table to get the correct angle for the cut.

You only need to measure the two angles before rotating the part. This is an exact solution.

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I don't know how I'd go about visualising that set-up - but I bet the first few I made would come out wrong

Practice with a cheap, easy to cut material.

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Thanks very much Paul. Great drawings and clearly explained. I was getting close to the solution in between baby sitting the last couple of days.

Thanks again ,John.

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With all due regard for the details above:
First you need the sine vise mounted on a rotary table to get the
three angles. :-)
...lew...

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Originally Posted by Lew Hartswick
With all due regard for the details above:
First you need the sine vise mounted on a rotary table to get the
three angles. :-)
...lew...
No, actually you don't. Just mounting the part at the first (vertical) angle in the vise jaws and then rotating the vise by the angle shown by my last equation will do it.

There may be three angles on the drawings, but any two of them will define the third so two settings are all that is needed. These angles can be easily set with angle gauges or protractor, depending on the accuracy needed. And, as I said, the solution is an exact one. If several tries are needed, it will be be due to an error in execution.

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"...but I bet the first few I made would come out wrong"

How would you know they're wrong?

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