I’m battling with something… I normally put down an arc at the waist line and then place a point on the arc at a certain distance from the centre of the arc for each leg of the waist dart. And the arcs become unsightly because I need to create a wide arc to cover a wide range of sizes.

My problem is determining the start & end of the arc so that it will increase or decrease according to size changes. Is there a way, using the functions and formulas, that I can calculate which degree to start with and end with, so that my arc is the minimun length required?

I hope I understand your problem a bit although I do not get the point of this construction.
What u try to achieve will not work. The segment of the arc will be always longer as the hypotenuse of the triangle. Actually u have to calculate the distance of the two points on the arc.

@JCDesign A28 is the original point on the arc. A41 is the new point using the sin() function so that I can eliminate using the arc that is sometimes too short. A41 should go on top of A28 to be accurate.

I’m trying to make an Isosceles triangle from 2 lengths. My algebra & trig are non-existant, so I have to browse the internet to find solutions and came across a YouTube video that showed that the formula is sin-1(3.7/12.6)=17.07, 17.07*2=34.14 (therefore 34.15).

Since yesterday, I have found asin whiich is the sin-1. However, I still don’t get to the result that he does in the video, even using his dimensions.

Bleh! ok, the reason why my results are different is because I’m using it wrong. This won’t work for what I want. (However, I’d still like to know how to do the calculation so that my result is the same as in the video.)

Ok, so back to the 1st question… How can I dynamically format the arc so that it is always just slightly larger than I need it to be?

All the sin cos etc functions are working only in a triangle with a 90°angle. Solution for a isolaticios triangle to split it in half. Means from Center a21 to 41 a line to angle at waist line.

asin works in radians, you want asinD for an answer in degrees. Unfortunately muparser, Seamly’s math parsing library, was coded by complete nerds (technical term for the sort of people who favor radians ) while most of us favor degrees. & so we have a program that expects an answer in degrees while defaulting to working your formula in radians.

Baby steps, @Douglas. I think I should start a topic where we pick a function and use it in a specific scenario in a pattern so that I can learn how to use them all.

Can’t wait to see the tutorial explaining the use of the hyperbolic cosine in patternmaking

Actually I found one while writing this: the hyperbolic cosine is the shape a thread makes when it is tied to something on both ends and is loose in between

But the hyperbolic functions have no simple “real world” definition, at least none that I’m aware of.

Let’s take the example of the hyperbolic sine, it is defined as the “odd part of the exponential function”, ie:

The only “graphical explanation” i know of is the one using the hyperbolic angle, but it is not “simple”

I can try to explain it in two words.

If you draw a unit circle, take a point on it, you’ll find the “classic” trigonometry functions by looking at the coordinates of the point, which are cos(t) and sin(t) where t is the angle, in radians, between the positive x-axis and the point:

The tangent can be found tracing… the tangent to the circle passing through the point (1,0). I can show it to you if you’re interested.

It is interesting to note (you’ll see why below) that t is also 2 times the blue area below:

Now for the hyperbolic functions. Instead of the unit circle, we need to draw a… unit hyperbola (in red below):

The coordinates of the point P are cosh(t) and sinh(t), where t is the hyperbolic angle. It is NOT the angle (Q,O,P), but it’s… 2 times the blue area.

So, to recap, it you draw a unit circle and take a point on it, 2 times the blue area gives the angle in radians and the coordinates of the point are cos and sin, and if you draw a unit hyperbola, 2 times the blue area gives the hyperbolic angle and the coordinates of the point are cosh and sinh.

I found the above illustrations on this website, it is more detailed and gives the proof of what I explained.

What I was trying to explain is that when you hold a cable, a chain or a thread by both ends, the shape it makes due to gravity is an hyperbolic cosine:

Ok, sooo… Using the last picture… If I was creating an armhole, let’s pretend that it’s the string on the 1st half of the image… And that the thread is a little heavy, so that the top attachment is at 90°.

How would I calculate the area to determine the curve of the string?

oooo0oooo

Ok, let’s use a real life example…

Line A9 to x7 is 2.5cm and is at 45°. How will I determine the length of the line A9 to x7a without creating the point?

If you know that A9-x7 is at 45° and is 2.5cm long then no matter the curve you need no other information to compute A9-x7a:

Let’s use new names:

A9 is A, x7 is B and x7a is C, with \alpha the 45° angle .

We have \cos\alpha=\frac{AC}{AB} then AC=AB\cos\alpha.

So, in our case: AC=2.5\cos45°=2.5*\frac{\sqrt{2}}{2}=1.77cm

Thus A9-x7a is 1.77cm.

Complex answer

But, if you want to use the fact that you know that the curve is a heavy thread, the question would be: what is the distance A9-x7a, given the fact that A9-x7 is at 45° ?

In that case we cannot say that A9-x7 is 2.5cm, we cannot chose this distance, it is imposed by the shape of the thread.

I’m not sure about the usefulness of answering this in patternmaking, but just for fun, let’s compute that

Schema

Let’s say that A15 is (0,0).

Curve equation

If the curve A10-x7-A15 is the first half of a heavy thread which is tied to A10 and to another point on the other side of A15, at the same height, like that:

And let’s say the thread is “perfect” and inextensible, of total length L, that is to say that the length of the curve A10-x7-A15 is L/2.

The equation of the thread is then y=a\cosh(\frac{x}{a}) , with a=\frac{0.25L²-h²}{2h} where h is the distance A10-A9.

(to prove those formulas we can split the thread in very small theoretical pieces, and then compute the forces applied to the pieces, we’ll just admit the formulas here…)

Equation of the line A9-x7

First, let’s compute the equation of the line A9-x7:

The equation of a line is y=bx+c.

Since the line is 45° we know that when x is increased by 1, y is also increased by one, thus b=1.

We know that the line pass through A9, thus through the point (-dA9, 0) where dA9 is the distance A9-A15.

So, using the above equation we have 0 = b*(-dA9)+c, thus c=dA9.

So the equation of the line is y=x+dA9

Equation of the intersection point x7

x7 is the intersection point of the line and the curve, so its coordinates (x,y) verifie both equations: y=x+dA9 and y=a\cosh(\frac{x}{a}).

Thus x+dA9 = a\cosh(\frac{x}{a}) and x = a\cosh(\frac{x}{a}) - dA9

Which is not easily solvable by hand if I’m not mistaken, so it should be solved using a computer or a calculator, so it’s not computable directly in Seamly…

And, if we get x, since we said that (0,0) is A15 we know that the distance x7a-A15 is minus the x we just computed, and thus we can compute A9-x7a, being the distance A9-A15 minus x7a-A15: