3/4/5 = Square
... Does not apply 😋🤣😆
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I guess that is what happens when there is a shortage of levels and framing squares in that part of the world.
And I'm guessing clear plastic tube and water to boot.
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Could be in Italy where everything is WOP sided.
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Howard, Fine & Howard General Contractors 😆
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At least the rain'll run off... somewhere.
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contractors of the leaning tower
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They probably used modern math.
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Built to specs - designed by Picasso!
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3 Stooges contracting.
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I've used and taught that method a lot. It's one thing I made sure all of my children understood.
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Shortage of levels.
Surplus of wine.
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When I took geometry an object with 3 sides that were 3 units long, 4 units long, and 5 units long was called a triangle, not a square. Must be non-Euclidian.
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Mr. Perfect: 30932649777563/comments/30932668226203
When I took geometry an object with 3 sides that were 3 units long, 4 units long, and 5 units long was called a triangle, not a square. Must be non-Euclidian.
The 3:4:5 triangle is the best way I know to determine with absolutely certainty that an angle is 90 degrees. This rule says that if one side of a triangle measures 3 and the adjacent side measures 4, then the diagonal between those two points must measure 5 in order for it to be a right triangle.
A SQUARE triangle = Right triangle
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Hmmmmmmmm. Are you sure?
In the Poincare' model, consider a Poincare' right triangle OAB, where O' = (0, 0) is the origin, and A = (x, 0) and B = (0, y) are points on the x- and y axes. If the angles at A and B are α and β, respectively, let δ = π/2 − α − β be the “defect” of the triangle. Then tan(δ/2) = xy.
The Poincare' line AB' is part of a circle Δ orthogonal to Γ. Let its center be C. Join CA, CB, and drop perpendiculars CD, CE to the two axes. Then ےACD = α and ےBCE = β, so ےACB = δ. Let A' be the other intersection of Δ with the x-axis. Since Δ is orthogonal to Γ, A' is the circular inverse of A, namely, A' = (1/x, 0). Since the arc AB subtends an angle δ at C, it subtends the angle δ/2 at A'. Thus tan(δ/2) = OB/OA' = y/(1/x) = xy.
"Ah hah!!!" you say. Not so fast, buster. If we translate this into a formula intrinsic to hyperbolic geometry, then we obtain the formula tan(δ/2) = cos F(a/2) cos F(b/2) given by Lobachevskii in his first publication on non-Euclidean geometry. Here δ is the area of the right triangle with legs a, b ..... and F, in Lobachevskii’s notation, is the angle of parallelism of a given segment. So neener neener neener. I get to use it.
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Nanuq907: 30932649777563/comments/30932683732251
Hmmmmmmmm. Are you sure?
https://us.v-cdn.net/6031683/uploads/81HY3QJQ2I2E/image.png
In the Poincare' model, consider a Poincare' right triangle OAB, where O' = (0, 0) is the origin, and A = (x, 0) and B = (0, y) are points on the x- and y axes. If the angles at A and B are α and β, respectively, let δ = π/2 − α − β be the “defect” of the triangle. Then tan(δ/2) = xy.
The Poincare' line AB' is part of a circle Δ orthogonal to Γ. Let its center be C. Join CA, CB, and drop perpendiculars CD, CE to the two axes. Then ےACD = α and ےBCE = β, so ےACB = δ. Let A' be the other intersection of Δ with the x-axis. Since Δ is orthogonal to Γ, A' is the circular inverse of A, namely, A' = (1/x, 0). Since the arc AB subtends an angle δ at C, it subtends the angle δ/2 at A'. Thus tan(δ/2) = OB/OA' = y/(1/x) = xy.
"Ah hah!!!" you say. Not so fast, buster. If we translate this into a formula intrinsic to hyperbolic geometry, then we obtain the formula tan(δ/2) = cos F(a/2) cos F(b/2) given by Lobachevskii in his first publication on non-Euclidean geometry. Here δ is the area of the right triangle with legs a, b ..... and F, in Lobachevskii’s notation, is the angle of parallelism of a given segment. So neener neener neener. I get to use it.
Section AB is an arc, not a line.
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Yes, but read on: since the arc AB subtends an angle δ at C, it subtends the angle δ/2 at A'. Thus tan(δ/2) = OB/OA'. We dropped perpendiculars.
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A squared + B squared = C squared Make the triangle proportional to 3/4/5 triangle and go as big as you want.
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I used this installing material, very handy when the roof, wall etc. is out of square. Of course use it to lay out ground projects or where ever it's needed. One starting line and you're good to go.
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Since we are are having fun:
For larger squares -
6/8/10 & 12/16/20
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yep, used it a lot.
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You guys are hurting my brain...Can we just discuss...Orthogonal Frequency Division Multiplexing or OFDM is a modulation format that is being used for many of the latest wireless and telecommunications standards??? 🤣
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Nanuq907: 30932649777563/comments/30932636314651
Yes, but read on: since the arc AB subtends an angle δ at C, it subtends the angle δ/2 at A'. Thus tan(δ/2) = OB/OA'. We dropped perpendiculars.
Indeed, and as I mentioned above... nonEuclidian.
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I tried OFDM and it made my corn flakes soggy. So that’s out.
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Used the ol' 3/4/5 many times doing tile and building layouts. Used it to check my brand new Dewalt 3 beam laser level and discovered it was off by 3/32" in 20'. Now that might not sound like much, but when you're laying out for over two thousand sq. ft. of 2"x2" tiles going around a swimming pool with runs of over 50' it adds up quick! Called 'em up and explained what was going on and had another one shipped on their dime the next day. I was impressed with their service.
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