# Thread: Division of a Length

1. ## Division of a Length

Need a Formula for calculating:

A length, X , is to be divided into a number of segments Y, the first segment the largest and each subsquent segment smaller in a linear regression so that the total of each segment size equals the length X.

Eample: X = 1 inch and Y=11 segments

Segment 1 > Segment 2, Segment 2 > Segment 3, ...............Segment 10 > Segment 11

The size reduction must be linear and when all segments are totaled the value equals X

2. n174k,
There are many ways to calculate this and more info is needed to solve the equation:

your average segment length will be =X/Y

is there a minimum length for segment 1?

Is there a specific angle that these segements must follow?

Will the segments be assembled to another component?

if so, do they need to be equally spaced?

Best regards

Bruno

3. n174k,
There are many ways to calculate this and more info is needed to solve the equation:

your average segment length will be =X/Y

is there a minimum length for segment 1?

No

Is there a specific angle that these segements must follow?

No

Will the segments be assembled to another component?

No

if so, do they need to be equally spaced?

No

Best regards

Bruno

What I am trying to accomplish is to include a formula in a threading macro where th total depth of thread = X and the number of roughing passes = Y.
Calculating equal depth of cut for each pass is easy, X/Y. I want the first past to be the greatest depth of cut ever decreasing in subsequent passes as the tool is removing more material as it reaches the final depth, minor diameter of an OD thread. The part program will only have the depth of thread, X and the number of roughing passes Y. X and Y will vary based on material and pitch of thread to be cut.

Thanks.

Del

4. Dear Del,

Thank-you for the thread. I never was that much good at maths, but here is my two cents...no, it is not a solution!

I think that you are talking about an "arithmetic series". I think this means that each member of the series is a fixed increment greater than the one before (or less, depending on which end you start).

Suppose you have Y members in the series, and that the total length of all members of the series is T.

I'm pretty sure that , as Bruno posted, the average length of all members of the series is T/Y.

Now, just suppose that member one has a length of L1, and member Y has a length of Ly. There are of course a total of Y members.

The average length of all the members will be ..

(L1+Ly)/2

So ...

(L1+Ly)/2 must equal T/Y

It is this precise point that I just give up!

Thanks for the challenge anyway..

Best wishes,

Martin

5. Keep working it Marty! Dust off that old textbook.

cheers

6. Originally Posted by cam1
Keep working it Marty! Dust off that old textbook.

cheers
Dear cam1,

Thank-you for your kind words of encouragement, but actually, I have reached the limit of my pathetic knowledge in these matters. I was always a bit "sub-prime" at maths, and whole libraries full of dusty text books will not help me now.

Best wishes,

Martin

7. I thought that I was bad in math, just had myself convinced. It was my Algebra that held me back. You know, multiply through by (1-X^2) to simplify the expression.
Once I solved that, the rest of math was an exercise in following and memorizing the rules.
Now I use my math to help test Gas Turbine Engines. That statement may want to make you reconsider flying in a Boeing Dreamliner. Not because it's a bad product, but if I turn ed out to be wrong with the math on the Engine mounts........

cheers

8. Originally Posted by cam1
..........with the math on the Engine mounts........

cheers
Dear cam1,

LOL. Henceforth, I shall only travel by ship...

Best wishes,

Martin

9. Wise choice, relax, getting there is the vacation. Very efficient stress free way to travel.
When or if aerospace tanks, perhaps I'll consider the shipbuilding industry they have machine shops on board.
Now...on with the math. We need results chop chop.

10. couldn't you start from the last pass and work your way back?
say you wanted the last pass to be .0005" deep. Then the previous pass could be twice that, the one before that double again. That way you would start out with passes of .01-.02 and work your way down to the final pass.

This might not be the most efficient way to go but it would work. If you want to get in depth you need to calculate the area of the cutter which is engaged. On the first pass, it would be a triangle with a height of the depth of cut. The next would be the triangle, minus the previous triangle, to give you the area of the cutter which was engaged.

Clear as mud?

I would help more but don't have time right now.

Good Luck
Matt

11. (My first post on CNCzone. Please go easy!)

If you work the problem backwards and assume that your last step is "1 unit", then the step before that would be "2 units", and so on. The first step would have been "Y units". Well, how long is "1 unit", then?

Well, the length of one unit would be the total length divided by the total number of units, which is (1 + 2 + 3 + .. + Y). What's (1 + 2 + 3 + ... + Y)? It's ((2 * Y) + 1) / 2.

So, after putting X back into the equation and doing some simplifying, "1 unit" would be (2 * X) / ((2 * Y) + 1). Let's just call that value "Z".

When actually machining, your first step would be Y * Z, the next would be (Y - 1) * Z, and so on down to 1 * Z.

12. Whoa! After reading my own post I realized I typed in an equation wrong. Definitely not good for a first post.

1 + 2 + 3 + ... + Y = Y * (Y + 1) / 2, not (2 * Y + 1) / 2.

So, fixing things up, the length Z would be (2 * X) / (Y * (Y + 1)).

Sorry 'bout that!

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