Af’noon all, sorry for the headache I’m about to create.

I’m stuck on a geometry issue at work. I have a right-angled triangle, opp & adj are 37.2mm long, hyp is 52.5mm. Now imagine the opp & adj continue on long lines, so the triangle is basically in the corner of a large square.

Now to throw a spanner in the works…. Place a circle on the triangle that will have it’s west and south points at the top and right side of the triangle. How can I find the radius of the circle when this is the only information available? I’m hoping there’s a formula for something like this!

See image for reference.

JTQ said:

Af’noon all, sorry for the headache I’m about to create.

I’m stuck on a geometry issue at work. I have a right-angled triangle, opp & adj are 37.2mm long, hyp is 52.5mm. Now imagine the opp & adj continue on long lines, so the triangle is basically in the corner of a large square.

Now to throw a spanner in the works…. Place a circle on the triangle that will have it’s west and south points at the top and right side of the triangle. How can I find the radius of the circle when this is the only information available? I’m hoping there’s a formula for something like this!

See image for reference.

Radius = 37.2 mm

JTQ said:

Af’noon all, sorry for the headache I’m about to create.

I’m stuck on a geometry issue at work. I have a right-angled triangle, opp & adj are 37.2mm long, hyp is 52.5mm. Now imagine the opp & adj continue on long lines, so the triangle is basically in the corner of a large square.

Now to throw a spanner in the works…. Place a circle on the triangle that will have it’s west and south points at the top and right side of the triangle. How can I find the radius of the circle when this is the only information available? I’m hoping there’s a formula for something like this!

See image for reference.

Draw perpendiculars to the two tangent points.

They will both pass through the centre, so their intersection point is the centre.

Does that answer the question?

mollwollfumble said:

JTQ said:

Af’noon all, sorry for the headache I’m about to create.

I’m stuck on a geometry issue at work. I have a right-angled triangle, opp & adj are 37.2mm long, hyp is 52.5mm. Now imagine the opp & adj continue on long lines, so the triangle is basically in the corner of a large square.

Now to throw a spanner in the works…. Place a circle on the triangle that will have it’s west and south points at the top and right side of the triangle. How can I find the radius of the circle when this is the only information available? I’m hoping there’s a formula for something like this!

See image for reference.

Radius = 37.2 mm

I was 27 seconds too slow!

JTQ said:

Af’noon all, sorry for the headache I’m about to create.

I’m stuck on a geometry issue at work. I have a right-angled triangle, opp & adj are 37.2mm long, hyp is 52.5mm. Now imagine the opp & adj continue on long lines, so the triangle is basically in the corner of a large square.

Now to throw a spanner in the works…. Place a circle on the triangle that will have it’s west and south points at the top and right side of the triangle. How can I find the radius of the circle when this is the only information available? I’m hoping there’s a formula for something like this!

See image for reference.

I think you’ve already found it Mr Q. It’s 37.2mm

lol… I realise my mistake after mollwollfumble gave the answer… obviously it’s 37.2mm. Sorry all, that’s the dumbest question I’ve asked today.

The Rev Dodgson said:

mollwollfumble said:

JTQ said:

Af’noon all, sorry for the headache I’m about to create.

I’m stuck on a geometry issue at work. I have a right-angled triangle, opp & adj are 37.2mm long, hyp is 52.5mm. Now imagine the opp & adj continue on long lines, so the triangle is basically in the corner of a large square.

Now to throw a spanner in the works…. Place a circle on the triangle that will have it’s west and south points at the top and right side of the triangle. How can I find the radius of the circle when this is the only information available? I’m hoping there’s a formula for something like this!

See image for reference.

Radius = 37.2 mm

I was 27 seconds too slow!

That was my third post, the wifi flaked out during the first two.

In Donde world it would depend on what value you use for the pi.
Now if you’ve got a line with a variable length of 9 subtending an arc…………….

JTQ said:

lol… I realise my mistake after mollwollfumble gave the answer… obviously it’s 37.2mm. Sorry all, that’s the dumbest question I’ve asked today.

No dumb questions only dumb people :P

Well this is the other half of what I’m working on… Same idea, just need to find the radius of the red circle, when only the triangle size is available.

Peak Warming Man said:

In Donde world it would depend on what value you use for the pi.

Yes. Thagoras would be most unhappy if you didn’t get the pi bit right.

Cymek said:

JTQ said:

lol… I realise my mistake after mollwollfumble gave the answer… obviously it’s 37.2mm. Sorry all, that’s the dumbest question I’ve asked today.

No dumb questions only dumb people :P

Yes. Better to keep one’s mouth shut, and be thought of as dumb, rather than open it and remove all doubt.

Pi equals 3 everyone knows that

• Well this is the other half of what I’m working on… Same idea, just need to find the radius of the red circle, when only the triangle size is available.

Circular segment

furious said:

• Well this is the other half of what I’m working on… Same idea, just need to find the radius of the red circle, when only the triangle size is available.

Circular segment

Thanks furious.

I was about to work through a more convoluted calculation, involving trigonometry.

2027.4

Peak Warming Man said:

2027.4

Damn.

He got it right.

is this cooked?

furious said:

• Well this is the other half of what I’m working on… Same idea, just need to find the radius of the red circle, when only the triangle size is available.

Circular segment

Thanks! :D And yep, 2027.4. Now I have the formula for that, I can get it sorted for work. Basically just creating a formula-driven table that works in our software to allow a cabinetmaker to put cut-outs for bins in his set of drawers.

JTQ said:

furious said:

• Well this is the other half of what I’m working on… Same idea, just need to find the radius of the red circle, when only the triangle size is available.

Circular segment

Thanks! :D And yep, 2027.4. Now I have the formula for that, I can get it sorted for work. Basically just creating a formula-driven table that works in our software to allow a cabinetmaker to put cut-outs for bins in his set of drawers.

That’s a pretty rubbish idea, I’d bin it before someone trashes it

Peak Warming Man said:

In Donde world it would depend on what value you use for the pi.
Now if you’ve got a line with a variable length of 9 subtending an arc…………….

Bugger.. you stole my thunder, But isn’t the Donde Pi exactly 3 if i recall to give him (or her) credit, Not just some weird random number.. that would just be stupid.

gaghalfrunt said:

Peak Warming Man said:

In Donde world it would depend on what value you use for the pi.
Now if you’ve got a line with a variable length of 9 subtending an arc…………….

Bugger.. you stole my thunder, But isn’t the Donde Pi exactly 3 if i recall to give him (or her) credit, Not just some weird random number.. that would just be stupid.

Mollwollfumble lives in Pi dimensional space.

mollwollfumble said:

gaghalfrunt said:

Peak Warming Man said:

In Donde world it would depend on what value you use for the pi.
Now if you’ve got a line with a variable length of 9 subtending an arc…………….

Bugger.. you stole my thunder, But isn’t the Donde Pi exactly 3 if i recall to give him (or her) credit, Not just some weird random number.. that would just be stupid.

Mollwollfumble lives in Pi dimensional space.

Have you seen my proof that Pi ~ 6?

mollwollfumble said:

mollwollfumble said:

gaghalfrunt said:

Bugger.. you stole my thunder, But isn’t the Donde Pi exactly 3 if i recall to give him (or her) credit, Not just some weird random number.. that would just be stupid.

Mollwollfumble lives in Pi dimensional space.

Have you seen my proof that Pi ~ 6?

No. Please show working.

Michael V said:

mollwollfumble said:

mollwollfumble said:

Mollwollfumble lives in Pi dimensional space.

Have you seen my proof that Pi ~ 6?

No. Please show working.

RND = RND

dv said:

Michael V said:

mollwollfumble said:

Have you seen my proof that Pi ~ 6?

No. Please show working.

RND = RND

I don’t get it. Please expand.

Michael V said:

dv said:

Michael V said:

No. Please show working.

RND = RND

I don’t get it. Please expand.

formatting fucked up

Basically if you round 6 to the nearest 1000, and round pi to the nearest 1000, you get the same answer, so they must be approximately the same.

dv said:

Michael V said:

dv said:

RND = RND

I don’t get it. Please expand.

formatting fucked up

Basically if you round 6 to the nearest 1000, and round pi to the nearest 1000, you get the same answer, so they must be approximately the same.

Oh, OK.

giggle

So the theory of nines was correct after all…

Michael V said:

mollwollfumble said:

mollwollfumble said:

Mollwollfumble lives in Pi dimensional space.

Have you seen my proof that Pi ~ 6?

No. Please show working.

A sphere has approximately the same volume as the cubic box that contains it.

So

4/3 pi r ³ ~ (2r) ³

So

4/3 pi ~ 8

So pi ~ 8*3/4 = 6.

(4/3 pi r 3 ~ (2r) 3 )

Yes I’m having a bit of difficulty on that first step. On what basis is this claim made?

dv said:

(4/3 pi r ³ ~ (2r) ³ )

Yes I’m having a bit of difficulty on that first step. On what basis is this claim made?

On the grounds that a sphere is tangent at all contact points to the cubic box that contains it. So the sphere must be similar in volume to the cubic box that contains it.

Much more similar in volume to that than it is to the volume of the largest cubic box that will fit inside the sphere, for example.

The cubic box has side equal to the diameter of the sphere, i.e. side 2r so has volume 2r cubed.

“On the grounds that a sphere is tangent at all contact points to the cubic box that contains it. So the sphere must be similar in volume to the cubic box that contains it.”

Well it must be less than the volume of the cubic box than contains it. I can give you a shape that’s like 1% of the volume of a cubic box that contains it.

mollwollfumble said:

Michael V said:

mollwollfumble said:

Have you seen my proof that Pi ~ 6?

No. Please show working.

A sphere has approximately the same volume as the cubic box that contains it.

So

4/3 pi r ³ ~ (2r) ³

So

4/3 pi ~ 8

So pi ~ 8*3/4 = 6.

A sphere has approximately the same volume as the cubic box that contains it.

Except that the cubic box is nearly double the volume of the sphere.

dv said:

“On the grounds that a sphere is tangent at all contact points to the cubic box that contains it. So the sphere must be similar in volume to the cubic box that contains it.”

Well it must be less than the volume of the cubic box than contains it. I can give you a shape that’s like 1% of the volume of a cubic box that contains it.

Try it. Give me a convex shape tangent to all six sides of the box that has a volume that’s like 1% of the volume of the box.

It can be done.

But if I was to say “give me a convex shape touching the centres of all six sides of the box that has a volume that’s like 1% of the volume of the box” then it can’t be done.

mollwollfumble said:

dv said:

“On the grounds that a sphere is tangent at all contact points to the cubic box that contains it. So the sphere must be similar in volume to the cubic box that contains it.”

Well it must be less than the volume of the cubic box than contains it. I can give you a shape that’s like 1% of the volume of a cubic box that contains it.

Try it. Give me a convex shape tangent to all six sides of the box that has a volume that’s like 1% of the volume of the box.

It can be done.

But if I was to say “give me a convex shape touching the centres of all six sides of the box that has a volume that’s like 1% of the volume of the box” then it can’t be done.

The criterion of convexity was not introduced in your original thesis.

dv said:

mollwollfumble said:

dv said:

“On the grounds that a sphere is tangent at all contact points to the cubic box that contains it. So the sphere must be similar in volume to the cubic box that contains it.”

Well it must be less than the volume of the cubic box than contains it. I can give you a shape that’s like 1% of the volume of a cubic box that contains it.

Try it. Give me a convex shape tangent to all six sides of the box that has a volume that’s like 1% of the volume of the box.

It can be done.

But if I was to say “give me a convex shape touching the centres of all six sides of the box that has a volume that’s like 1% of the volume of the box” then it can’t be done.

The criterion of convexity was not introduced in your original thesis.

Well, duh, a sphere is convex.

mollwollfumble said:

dv said:

mollwollfumble said:

Try it. Give me a convex shape tangent to all six sides of the box that has a volume that’s like 1% of the volume of the box.

It can be done.

But if I was to say “give me a convex shape touching the centres of all six sides of the box that has a volume that’s like 1% of the volume of the box” then it can’t be done.

The criterion of convexity was not introduced in your original thesis.

Well, duh, a sphere is convex.

Well yes it is.