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Godspanther

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Godspanther

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Svein

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The arc is the essential information here. What does it mean to know an arc of the circle? An arc has obviously at least three points which are enough to determine the circle.All you can see is A, B the straight lineand an arc representing an unknown number of degrees.

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Lnewqban

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Godspanther

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Lnewqban

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That formula exists, it is used in construction.

Is this homework?

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If you have the arc length above the chord, then we are left with two possibilities. If we have a third point, then we have the circle.

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hutchphd

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If one knows the sagitta then $$ r=\frac { s^2+l^2} {2s}$$ where ##l=AB/2##

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Godspanther

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Lol. No. I am 41 years old. I have a passing interest in forging. I am contemplating what would be the optimal degree of arc to create a perfect scimitar. The closest I can find online is this one. https://www.trueswords.com/Arabian-...vPlyZYonirFBQ-I4qoHK6Zjy3gk0mikuhmaxCVicPuhys But the blade is short. If I can find a formula I can print the picture, make the measurements and figure out the arc for a blade between 32-36 inches. I am still debating best blade length and will likely examine multiple designs.That formula exists, it is used in construction.

Is this homework?

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Lnewqban

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Very interesting project.Lol. No. I am 41 years old. I have a passing interest in forging. I am contemplating what would be the optimal degree of arc to create a perfect scimitar. The closest I can find online is this one. https://www.trueswords.com/Arabian-...vPlyZYonirFBQ-I4qoHK6Zjy3gk0mikuhmaxCVicPuhys But the blade is short. If I can find a formula I can print the picture, make the measurements and figure out the arc for a blade between 32-36 inches. I am still debating best blade length and will likely examine multiple designs.

Please, see:

https://www.mathsisfun.com/geometry/arc.html

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Godspanther

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That is just the sort of thing I was looking for. That you very much.

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Svein

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First: I did not get the fact that you also have got the arc itself, not just the line AB. Second: The formula I quoted above is mathematically correct, but when the height is small relative to the width, the uncertainty in the radius calculation can easily be very large.Very interesting project.

Please, see:

https://www.mathsisfun.com/geometry/arc.html

View attachment 298580

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hutchphd

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uncertainty in the radius calculation can easily be very large.

Yes when grinding a telescope mirror you use it to calculate the depth of the sagitta required to produce a mirror of desired radius ( and therefore focus) from a circular glass blank. Commonly a "feeler" gauge of some sort is usually created to test "s" the sag depth as you grind (you call it "height"). Because you know l=AB/2 very accurately for the mirror blank the technique is quite precise.

But if you measure l with error ##\Delta l## then$$ r=\frac { s^2+l^2} {2s}$$ $$\Delta r=\frac {\partial r } {\partial l} \Delta l$$ $$\Delta r=\frac l s \Delta l$$ As you have noted

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Lnewqban

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You are absolutely correct; the method is as accurate as the measurements.First: I did not get the fact that you also have got the arc itself, not just the line AB. Second: The formula I quoted above is mathematically correct, but when the height is small relative to the width, the uncertainty in the radius calculation can easily be very large.

As you have seen, the OP just wants a rough estimate so he can shape a sword.

I have seen it used in buildings construction as well, where errors are not that critical either.

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Svein

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hutchphd

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For completeness sake:

if the errors are independent $$(\Delta r)^2=|\frac{\partial r}{\partial s}|^2 (\Delta s)^2+|\frac{\partial r}{\partial l}|^2|\Delta l)^2$$

For me the cleanest way to write this is

$$(\Delta r)^2=\left |1-\frac r s\right |^2~ (\Delta s)^2+\left|\frac l s\right |^2(~\Delta l)^2$$

(Revised)

If I can find a formula I can print the picture, make the measurements and figure out the arc for a blade between 32-36 inches. I am still debating best blade length and will likely examine multiple designs.

If the sword is really going to be a circular segment (no modifications) wouldn't it be easier just to make a big compass drawing with a stick (or string) and some chalk on cardboard? Anyhow you have (re) learned some geometry. Have fun

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