[2_jammer]

Hello All,
I'm looking for some help on trying to figure out where these two radiuses meet from the datum. I can figure it out using my CAD program, after all, I drew it using CAD, but what I'm trying to do is figure it out long hand using trigonometry. I'm hoping to get a better grasp of programming by learning things "the hard way". I'd appreciate it if someone could steer me in the right way. Please see the attached file. Thanks, Shane.

[ger21]

The point your looking for is on a line connecting the center of the two arcs. The radius is the length of the hypotenuse of a right triangle. If you need more steering, I'll have to brush up on my trig.

[Kevin Taylor]

You know the arc center's and divise the stop and start point's by drawing triangles and doing hypontious theroy aneasy way to rember it is 3/4/5 3x3=9+4x4=16hens 9+16=25 sqroot is 5by imagining triangels by conecting the point's you can come up with your number's an old text book or I think the machinest handbook has a lot of trig info in it If you want a tool path in your bobcad go to other after you select your line and click on offset and chose right or left and the tool radis it will generate the tool line Good luck Kevin

[Geof]

I have to admit my trig is very rusty but I don't think you have enough information. You have the radii for the two curves and the position of the center of one curve on the X axis, or at least that is what I assume your 0.8 dimension is, but you do not have the position of the center of the other curve. I think you need the position of the two center points for a solution.

[2_jammer]

Thanks for the replies- this example is taken out of a book. According to the author, the center of the 2" radius can be calculated by the dimensions presented. I know I'm pretty rusty with my trig, but I just can't seem to figure out how to set this up. Ger 21, I guess I need a bit more steering...

[Geof]

I was too pessimistic, your book is correct. The 2.0000 radius circle is centered at 3.7206, 3.6078 and the tangent point between the circles is at 1.8952, 2.7904.

[ger21]

I was assuming you know the centers of the 2 arcs, is this correct?

[2_jammer]

Geof-
Could you explain to me how you came up with those cooridinates.

ger21- I know the center point of the 1.200 radius, I think I need to trig out the 2.000 radius?

Thanks once again for the help guys!

[ger21]

If you don't know the center of the 2" radius, I think you'll need more info to locate it.

[Geof]

The calculation uses the idea of similar triangles; that is triangles with the same angles but different length sides. I have drawn four extra lines on your picture; RED joins the centers of the two circles and passes through the tangent point. The length of this hypotenuse is the sum of the radii. BLUE travels along the X direction from the R1.2 center past the center of the other circle; YELLOW travels along the Y direction from the center of the R2.0 circle and intersects BLUE at a right angle. The length of YELLOW is the distance between the two centers in the Y direction. The length of BLUE is calculated using Pythagoras theorem. edit: This is the distance between the two centers in the x direction.

The PINK line from the tangent point down to BLUE forms another right angle triangle and this hypotenuse is 1.2 because it is the radius of the circle. The length of y can be calculated because y divided by 1.2 equals YELLOW divided by RED. Similarly the small section on the blue line labelled x can be calculated because x divided by 1.2 equals BLUE divided by RED.

The tangent point is at 0.8 + x and 2.3 + y.

[WhiteTiger]

Trig is based on the concept that all triangles have 6 elements (three angles and 3 side lengths). If you know any 3 elements you can solve for the other three. In layout work it's almost always done by deriving a right triangle, since that provides you with one known element, the 90 degree angle and leaves you with a requirement of only two other elements.

Y position of the right side arc center is known (2.0000 + 1.6078= 3.6078). Yposition of the left side arc is known (3.5000 - 1.2000= 2.3000). Yellow side length is then 3.6078 - 2.3000= 1.3078.

You now have hypotenuse (sum of radii) and one side of the large triangle known along with the 90 degree angle. Solve for the angle of the hypotenuse of the large triangle, then use that angle and hypotenuse of the similar small triangle as knowns to solve for x and y of the small triangle, and those will tell you what to add to x and y center of the small triangle to find your point of tangency.

EDIT= corrected wording of parenthetic note on hypotenuse of large triangle
Tiger

[Geof]

It is not necessary to solve for any angles.

[WhiteTiger]

True, but he did say he was doing it the hard way as an exercise. It can be done either way, just thought it might be beneficial to have the angular approach explained too


Tiger

[2_jammer]

:cheers:
Hey Guys-
Thanks for all your help!
Geof I appreciate your drawing, that was exactly what I needed to help visualize that point I was looking for.
White Tiger, Thanks for your input to with your explanation.

Can anyone recommend a good CNC programming book? I've been programming for a number of years via CAD/CAM, and I really never learned how to figure things out long-hand. (chair) Shane.

[lerman]

Trig is based on the concept that all triangles have 6 elements (three angles and 3 side lengths). If you know any 3 elements you can solve for the other three. In layout work it's almost always done by deriving a right triangle, since that provides you with one known element, the 90 degree angle and leaves you with a requirement of only two other elements.

Y position of the right side arc center is known (2.0000 + 1.6078= 3.6078). Yposition of the left side arc is known (3.5000 - 1.2000= 2.3000). Yellow side length is then 3.6078 - 2.3000= 1.3078.

You now have hypotenuse (sum of radii) and one side of the large triangle known along with the 90 degree angle. Solve for the angle of the hypotenuse of the large triangle, then use that angle and hypotenuse of the similar small triangle as knowns to solve for x and y of the small triangle, and those will tell you what to add to x and y center of the small triangle to find your point of tangency.

EDIT= corrected wording of parenthetic note on hypotenuse of large triangle
Tiger

It is not true that if you know any three elements of a triangle you can determine the other three. One of the three known elements must be a side.

Ken

[WhiteTiger]

LOL... I was wondering who was going to correct my wording on that one. Started to edit it a couple of times but kept getting interrupted by real life and not getting it done


Tiger

[Geof]

Can anyone recommend a good CNC programming book? I've been programming for a number of years via CAD/CAM, and I really never learned how to figure things out long-hand. (chair) Shane.

"CNC Programming Principles and Applications" by Mike Mattson, published by Delmar/Thomson Learning, 2001 is the best I have seen. I have only one criticism and that is the author only describes using G92 for creating supplementary work zeroes I much prefer G52. He covers the math; algebra, geometry and trigonometry very well and just to the correct depth I think.

[underdog]

Shane / 2 Jammer,

I really admire your wanting to learn the nuts and bolts of your craft. I once had that much energy too - and I can say I am better for having done things the hard way to see what makes them tick.

A really great CNC programmign book is "Programming of CNC machines" by Ken Evans. Available from Travers tool co 1-800-221-0270 for $37.00 (US). It is their item number 99-065-054.

Good luck.

[2_jammer]

Thanks Underdog- I appreciate your comment. I'll look at the books that you and Geof have suggested. Shane.

[lakeside]

well you must have some time on your hands to even want to lean that type of math but if your going to spend $37 on a book why not make it real Interesting and put the pocket calculator away and dust off the slide rule