geometry expert required

[bugbear]

Just curious, does anyone here actually own a bokken, or a katana for that matter?

I've made a few staff and short staffs for a mate who's into martial arts.

I used hickory.

BugBear

 

[MikeG.]

Damn you and your ruler that can measure 0.038 of an inch. :-D

That's about half way between a 16th and a 32nd. The ruler can do that. My eyes can't do that any more if I forget to take my glasses out to the workshop.

 

[sunnybob]

Just curious, does anyone here actually own a bokken, or a katana for that matter?

yeah, me.
just-for-fun-t110050.html

this is the mark 2 improved version.


Naz.... bizarre or what?
Not only do I know of occams razor, I OWN a gillette mark 3!

 

[DTR]

Damn you and your ruler that can measure 0.038 of an inch. :-D

That's about half way between a 16th and a 32nd. The ruler can do that. My eyes can't do that any more if I forget to take my glasses out to the workshop.

It's nigh on 1mm, if you can tolerate being 35 microns short...

 

[DTR]

Here's the maths IF and ONLY IF the curve you want is a circle. (If it isn't a circle, it is not defined).

the formula you want is

r^2 = a^2 + (r-s)^2

r is the radius of the circle
a is the half-length of the chord, 0.5 m in your case
s is the distance from the chord to the outside of the circle at the mid point, 0.02 m in your case.

Keith

Thanks for this, I've lost count of how often I could have used this formula. I really ought to write it down somewhere

 

[bushwhaker]

Length the curve must be 110.35 cm.

 

[MikeG.]

Length the curve must be 110.35 cm.

No, no, that's miles out. The length of the arc is 1001.1mm

 

[Sheffield Tony]

You don't really need to write this down. It is just Pythagoras theorem. Draw your arc, and the sector of the circle it forms part of. Draw a line bisecting the sector, and a straight line connecting the two end points of your arc. You now have a sort of kite shape with a rounded top. The two right angled triangles have hypotenuse r, adjacent (r-s) and opposite a.

 

[NazNomad]

The two right angled triangles have hypotenuse r, adjacent (r-s) and opposite a.

 

[bugbear]

You don't really need to write this down. It is just Pythagoras theorem. Draw your arc, and the sector of the circle it forms part of. Draw a line bisecting the sector, and a straight line connecting the two end points of your arc. You now have a sort of kite shape with a rounded top. The two right angled triangles have hypotenuse r, adjacent (r-s) and opposite a.

Measuring the sagitta for an existing arc is easy, as you say.

Working out the radius that will give a particular sagitta is what the equation is for.

BugBear

 

[Woody2Shoes]

Last time I used this was for copying an arch over a fireplace - easy to measure the span and the "rise" across that span, but hard to find the radius for the router jig.

Here's a nice website that shows you the maths and provides a calculator (the second one down) which matches MikeG's 6250 figure.

http://www.liutaiomottola.com/formulae/sag.htm


Hope no-one gets injured by your results!

 

[Sheffield Tony]

I'm not talking about measuring anything, by "draw" I meant "draw a diagram of". This is how to get to the equation to save the bother of remembering it.

This reminds me of my annoyance at my son's GCSE maths syllabus. Lots of methods and rules learned and applied laboriously, precious little of understanding of where they come from and why they work. Which is, of course, the important part.

 

[MikeG.]

Last time I used this was for copying an arch over a fireplace - easy to measure the span and the "rise" across that span, but hard to find the radius for the router jig.

Here's a nice website that shows you the maths and provides a calculator (the second one down) which matches MikeG's 6250 figure.

6260!!

 

[bushwhaker]

I think

So,
r = b/2 + a2/(8·b)
α = 2·asin(a/(2·r)) °
S = π·r·α/180

 

[sunnybob]

Hey, I've discovered perpetual motion.

It seems to have passed a lot of folks by, but my problem was solved on page two. ROFLMFAO.

 

[NazNomad]

... It seems to have passed a lot of folks by, but my problem was solved on page two.

Trigonometry, it's the gift that keeps on giving. 8)

 

[ColeyS1]

Hey, I've discovered perpetual motion.

It seems to have passed a lot of folks by, but my problem was solved on page two. ROFLMFAO.

Just because you've finished with the thread, doesn't mean it can't continue so everyone learns something [WINKING FACE]

Sent from my SM-G900F using Tapatalk

 

[sunnybob]

I just hope whoever is trying to learn something has a greater understanding (and attention span) than i have.

Carry on, dont mind me, i'll just sit in the corner looking wise. (g)

 

[bushwhaker]

but my problem was solved on page two. ROFLMFAO.

Mea culpa!

 

[sunnybob]

I just googled "yambol"

Hope i can still remember where it is tomorrow.

(My internet is weird and strange by turns, today I cant access the smiley page so assume there are a couple after every one of my posts)