Boohoo me. It's been years since I've had to do any math. I have a little math problem to solve but I'm stumped. I forgot too much about trigonometry. I've drawn up a sketch in Inkscape that illustrates the problem. Can someone calculate the angle a and the radius r? And preferably explain how you calculated it. Maybe it will jog my memory :-)
Background: I'm going to need to draw and saw out the grey area out of wood. The radius is the length of the rope I need to tie to my pencil. They are going to be the ribs of a shield press (for making medieval wooden curved shields).
Well, to get the radius, we can do (61/2)^2+(r-16.5)^2=r^2 (Pythagorean Theorem) which gives r = 36.4394. Now, if we visually rotate the circle to the right, we'll notice that sin(a) = (61/2)/r which gives us a = 56.83 degrees (we divide by the radius so that we're working with a unit circle, which is what sin uses)
Quote:Original post by nullsquared Now if we visually rotate the circle to the right, we'll notice that sin(a) = (61/2)/r which gives us a = 56.83 degrees.
I might be wrong, but shouldn't that be sin(a) = (61/2)/(r-16.5) ?
Quote:Original post by nullsquared Well, to get the radius, we can do (61/2)^2+(r-16.5)^2=r^2
Duh! Thanks! I've been thinking waaay to complicated, trying to figure out r from sine and cosine. Basically I ended up with a system with two unknowns (both a and r) that I couldn't solve.
Quote:Original post by nullsquared Now if we visually rotate the circle to the right, we'll notice that sin(a) = (61/2)/r which gives us a = 56.83 degrees.
I might be wrong, but shouldn't that be sin(a) = (61/2)/(r-16.5) ?
I don't believe so because we're working with the full circle (the line segment to the point (r cos(a), r sin(a)) extends r units). Maybe you're thinking of using cosine? In which case, we can say that cos(a) = (r-16.5)/r.
Firstly, computing the radius (r) would be your first point of call.
A simple application of Pythagoras' theory will suffice:
r^2 = (61/2)^2 + (r - 16.5)^2
This assumes of course that the internal triangle is right angled and that the chord is bisected into two equal parts by the vertically aligned radius.
Solving the theorem yields that the radius becomes:
r = 36.44
to 2 decimal places.
Computing the angle (a) now becomes trivial, since all sides of the internal right-angled triangle are known.
So using the tan (sin and cos could also be used!):
For making shield presses I've always found it worked better to just use a lightly rounded center bar, and then straps to pull it down. You are never going to get the finish shield to perfectly match your frame anyway, they always spring back a little. (Unless you're using fiberglass over a mold or something.)
You can refine your shape by using other blocks to either side of your center.
Old Username: Talroth
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Quote:Original post by Talroth For making shield presses I've always found it worked better to just use a lightly rounded center bar, and then straps to pull it down.
That works, but they end up somewhat V-shaped instead of evenly round. It doesn't matter if they spring back a little, but I want the shape to be evenly round.
