Suppose I have an archway, I know the width and the height of the arch. I know how to find the radius but I don’t know how to find what portion of the circumference my arch is?
Width divided by the diameter gives you the sin of the central angle.
If I’m picturing this right, the angle of the arch is 2 times the arcsine of half the width divided by the radius. The portion of the circumference is that angle divided by 2 pi.
I will have to study up on sin but I think I can get it from here. I wish I would have stayed in school a bit longer I never thought I would need this stuff.
By portion of the circumference, do you mean the angle subtended? Law of cosines:
cos[sup]-1[/sup] (1 - w[sup]2[/sup]/2r[sup]2[/sup])
On preview, I think Andy L calculated the same thing using half the subtended angle instead while I was searching for the vbulletin code for superscript.
What I am doing is some illustrations of bows and arrows. I know from experience that If I use a string 3" shorter than the bow I will get a braced height of about 6". I want to be a bit more precise in my illustrations as I will be using quite a few different radiuses.
So by portion of the circle I mean if I know the width and height I also want to know the exact length of the portion that is arched.I don’t know if this is a concern for you, but the shape of a bow is not generally an arc of a circle.
I will actually be working with portions of the limb that are bending in a radius. The english long bow is often tillered to bend in a smooth continuous arc, But I primarily work with stiff handled bows that do not. The particular thing that I will be demonstrating can be best illustrated and expalined using each limb tillered into a continuous radius.
As I understand it, you are looking for a calculation which will give you the length, presumably in inches, of your curved material? I’m not sure I know the technically precise way to write out the following equation (corrections welcome), but when I was attempting to determine the length of a curved landscape wall in linear feet, my surveyor gave me this formula:
[(°degree of angle x pi)÷180] x Radius
That would be the angle at which your two radii converge, and it presumes you have fixed points at which your curve begins and ends. For all I know, this may be the same thing someone more mathematically savvy has already suggested above – or may solve a different problem from the one you’re trying to work out! Thought I’d just throw it into the pot, on the off chance that it’s what you’ve been looking for.
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That really was helpful, gave me a better grasp on what I am looking at. My math stopped at arithmatic so anything with letters or symbols I have to research a bit to figure them out.
Not to be rude, but how did that happen? I thought everyone was required to take algebra (a couple of years worth), trigonometry (what your question is), and geometry at a bare minimum to graduate from high school.
We did anyway, with the added fun of a semester of computer programming and a year of pre-calculus, which is basically where algebra and trigonometry are combined without using derivatives or integrals.
I got a little taste of alegebra in the 8th grade but my lifestyle changed right about that point and my education pretty much came to an end. I was using the Khan academy online to advance my math skills but the problem I had with that was nothing seemed to stick for than a few days. I advanced really rapidly by following the direction and taking the tests but I found without some practical applicatian or a lot more practice I just retained all most nothing.
Once I saw the diagrams of how it was figured out I knew immediately how to do it but I did not know what sine, cosine, or the other symbols actually meant, I only had some idea what they meant. I went to high school in the 60's. I never carried a single book to a single class or did a single homework assignment and still somehow managed to graduate. I went to night school for the goverment and history classes I had to make up.