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View Full Version : hjaving brain fart, need formula

idacal
01-22-2012, 07:59 PM
my lathe is an oldy and every line on the indicators is 4 thousandths of stock removal, really hard to use for precision, I have been running the compound on a 5 degree taper to the work for stuff that needs to be close. but I havent been able to find the formula that I need: say I run the compound forward 10 thousandths how much stock removal at 5 degrees am I going to get? I had a trig book years ago but I cant find it so any formulas or help would be great thanks

idacal
01-22-2012, 08:01 PM
nothing like having a misspelling in the title sorry.

tapwelder
01-22-2012, 09:07 PM
.01sin5 will give you the distance traveled forward.

Oldiron2
01-22-2012, 10:16 PM
I don't have the tables handy so can't answer your specific question, but do know that if you set the compound to a hair over 84*, (90* being parallel to the ways), 10 units on the compound will move the bit inward 1 unit.
If your lathe has 0.002" movement per grad, you should get 0.0002" movement inward at this angle so 2.5 grads should give 0.001 removal.

Fred s
01-23-2012, 07:52 AM
Take the inverse sine (arcsine) of how much movement you want perpedicular to the ways divided by how much movement you want parallel to the compound feed

if you want .001 of perpendicular movement for every .012 of compound movement, it come out to a 4.78 deg angle. now all you have to decide is whether you want .001 off of the radius, or off of the diameter.

Bistineau
01-23-2012, 01:52 PM
nothing like having a misspelling in the title sorry.

That's a result of the brain fart, it's to be expected at times during one of those.:(

Warpspeed
01-24-2012, 04:03 AM
Fred has nailed it.

Job RADIUS reduction = Tool travel along taper slide x Sin 5.0 degrees

So 1.0 thou advance into the job, requires 11.47 thou on the taper slide.
Three of those .004" divisions would come mighty close to removing two thou from the work.

Oldiron2
01-24-2012, 02:25 PM
Fred has nailed it.

Job RADIUS reduction = Tool travel along taper slide x Sin 5.0 degrees

So 1.0 thou advance into the job, requires 11.47 thou on the taper slide.
Three of those .004" divisions would come mighty close to removing two thou from the work.

Just curious; why is one part in 11.47 better than one in 10.0? Seems to me that the one in ten is much easier to work with and to remember for next time.

Again, Just Curious.....

idacal
01-24-2012, 03:11 PM
thanks for all the help, I want it as simple as possible, I will just change the taper 84 degrees so its 1 in ten, again thanks for all the help

Fred s
01-24-2012, 03:57 PM
at first he said .004 per increment, so i picked .012 as a good way to stay on a line.

Hey, idacal just noticed you are from ID, I grew up way up north, CdA way

aevald
01-24-2012, 05:35 PM
Hello idacal, if I understood your issue correctly, I believe this jpg. might illustrate an approach that might work for your question. Best regards, Allan

rlitman
01-24-2012, 06:23 PM
Hello idacal, if I understood your issue correctly, I believe this jpg. might illustrate an approach that might work for your question. Best regards, Allan

Right, so if his compound and cross-slide are graduated in 0.002" increments (typical if you don't have "micrometer" dials), then at that angle, each line on the compound represents 0.001" of reduction in diameter.

aevald
01-24-2012, 06:48 PM
Hello rlitman, if I read his post correctly and understood it the way that I am visualizing it, I would agree with you. Regards, Allan

OPUS FERRO
02-03-2012, 10:43 PM
idacal - It will reduce your cutting tool to center travel, but if you turn
the compound to 45 degrees; the math should be - 2 to 1. Opus