How to measure a length from arc



So, I’m making shirt sleeve for menswear now and having some difficulties. I need to measure a length from an arc. Here’s the pic:

sleeve men val

To have the arc, firstly I use “arc” tool from crown 3 with given radius, for example 27cm. After that, I need to measure (for example) 24cm onto the center line from the previous arc. So I don’t know how long it is from crown 3 or any other point. Do you know how to do it?

Here is the how I want it to be:

sleeve men

I want to make new biceps line drawn in red. Here is my files if you need it:

menswear shirts.val (35.5 KB)

measurements menswear.vit (1.4 KB)

Please help. Thanks a lot!


Why do you need arc if you just need points of the right angle triangle,you can use trigonometry? The point of the central line you are looking for has the length from crown3 equal to sqrt(27^2-24^2) ( see Pythagorean theorem)


I am not sure whether I understand what you are looking for. If you want to know the length of the splines you can just look them up in the variables table; they also show when you hoover over the spline: Spline_Variables_Table




@Olgatron Sadly, I’m not good at math… I don’t understand what you’re saying…

@moniaqua No, I need to determine the new back sleeve corner at biceps level as in the picture with the red line. The instruction is 27cm from lowered sleeve cap diagonally, and 24 cm from the center line… I determine the 27cm using the arc at a distance tool, but the problem is, I need to measure 24cm from the 27cm-radius arc…


If you put a new point from Crown3 down along the central line) at the length equal to 12.37 cm and restore a perpendicular in this point equal to 24 cm you will get exactly the new back sleeve corner as per your image (and no ark needed)


Ah, ok. The arc doesn’t help you at all here. Proceeding as @Olgatron says could help :slight_smile: Math is not that bad; Pythagoras helps you to get the distance from crown3 to the third point of your triangle, as it has a 90° angle.

He says:


a^2 + b^2 = c^2 | - b^2

a^2 = c^2 - b^2 | sqrt

a = sqrt(c^2 - b^2)

So, if c = 27 cm and b = 24 cm, a = sqrt((27 cm) ^2 - (24 cm) ^2) = 12,37 cm, like Olga said :slight_smile:


@Olgatron @moniaqua

Ahh, I see! Thank you so much!

Btw, what’s wrong with this formula? Where’s the mistake?


fyi, ((Spl_slf4a1a2_chest5+Spl_slf5_chestfront4+Line_slf4_slf5)/2) is C and ((Line_chest_chest2+Line_chestfront_chestfront2)/2) is B

Edit: Don’t mind. The value are swapped


I didn’t have the brain yesterday to think it through,sorry. I am trying to understand what you want to calculate. Splines and Pythagoras don’t get to well together; Pythagoras ist for straight lines. Also, in

I see four opening brackets, but only three closing ones. Probably the fourth in the beginning leads to wrong calculation, but actually the editor should give an error message.



It’s okay, thank you for replying! Hmmm, say I want to calculate A. B is the front and back underarm length, and C is the front and back armhole length (thus the spline). I think I found the problem, that B is longer than C. I don’t know how that happen, is that mean that my pattern is wrong? Anyway, I swapped the value, the longer one minus the shorter one (B-C) and I found the value of A. Would there be a problem with me swapping the value?

Oh, for the bracket, the fourth bracket is at the end.