Forgot my geometry...

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I've got a board set at a 45 degree angle, back from a line. How much (percentage) of the length of the board does it take up? To conceptualize the issue, I drew a one inch line on paper with a ruler, and rotated the ruler to a 45 degree angle, thinking that the one inch mark on the ruler would be only 1/2 away from the starting point (along the original path of the ruler), but it looks like it's about 90% along the one inch span. What's the formula?
Dave
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Hope the ASCII art is legible...
.707" |---|
/ / / 45 deg angle /_______
|- 1" -|
The length of the diagonal line is 1".
The diagonal of a 1" square is 1.414"
--
JeffB
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Forget my last post. NOW i see my mistake: I eyeballed the 45 degree angle wrong--I had it a bit less than 45. It's as you said.
Now I can start cutting some wood! Thanks, Jeff
Dave
JeffB wrote:

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I'll admit, I'm having a little trouble following exactly what you are describing, but are you looking for the standard trig formulas?
Sine = Opposite/Hypotenuse Cosine = Adjacent/Hypotenuse Tangent = Opposite/Adjacent
If you know what your angle between the two 1" lines is, you should be able to get the appropriate angle value (Sin, Cos, or Tan) from a decent scientific calculator or a Trig table, and then just use standard algebra rules to solve for your missing dimention. Aut inveniam viam aut faciam
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If you mean a miter cut, the length of the miter is
the root of two times the square of the width of the board.
If you mean a bevel cut, the length of the bevel is
the root of two times the square of the thickness of the board.
1 inch wide = 1.4142135623730950488016887242097 2 inch wide = 2.8284271247461900976033774484194 3 inch wide = 4.24264068711928514640506617262909 4 inch wide = 5.65685424949238019520675489683879
It appears the bevel/miter is proportional to the width by a factor of ~1.41. Or, the width/thickness is always 70.7106781186547524400844362105198% of the bevel/miter.
--
PDQ --
| I've got a board set at a 45 degree angle, back from a line. How much | (percentage) of the length of the board does it take up? To | conceptualize the issue, I drew a one inch line on paper with a ruler, | and rotated the ruler to a 45 degree angle, thinking that the one inch | mark on the ruler would be only 1/2 away from the starting point (along | the original path of the ruler), but it looks like it's about 90% along | the one inch span. What's the formula? | | Dave
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Try again. Square root of 2 times the width of the board _not_ squared.

Try again. Square root of 2 times the thickness of the board _not_ squared. Your formulas below are correct (even though given with an absurd degree of precision), but your descriptions above are wrong, and don't match the formulas.

Yes. Proportional to the width. Not to the square of the width.

70.7 % is plenty close enough.
-- Regards, Doug Miller (alphageek at milmac dot com)
Nobody ever left footprints in the sands of time by sitting on his butt. And who wants to leave buttprints in the sands of time?
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Picky, picky, picky.
If you want to play those games, Doug:
"Bevel" is described as "the angle formed at the juncture of two non perpendicular surfaces."
"Miter" could mean "a tall ornamental liturgical headdress" worn by some members of the clergy, or it could mean, as it does in this case, "either of the surfaces that come together in a miter joint".
If you want to play with polygonal surfaces, why not say so? "board _not_ squared" is so imprecise.
I guess your problem must lie with your inability to visualize the position of the board within its frame of reference.
I am further amazed that one who would advertise one's self as a "Geek" would be unable to appreciate the intended absurdity of the precision. I was leaving it up the positor, to extract a suitable level of imprecision.
Go play with your semantics, sirrah.
--
PDQ --
| >If you mean a miter cut, the length of the miter is | > | >the root of two times the square of the width of the board. | | Try again. Square root of 2 times the width of the board _not_ squared. | > | >If you mean a bevel cut, the length of the bevel is | > | >the root of two times the square of the thickness of the board. | | Try again. Square root of 2 times the thickness of the board _not_ squared. | Your formulas below are correct (even though given with an absurd degree of | precision), but your descriptions above are wrong, and don't match the | formulas. | > | >1 inch wide = 1.4142135623730950488016887242097 | >2 inch wide = 2.8284271247461900976033774484194 | >3 inch wide = 4.24264068711928514640506617262909 | >4 inch wide = 5.65685424949238019520675489683879 | > | >It appears the bevel/miter is proportional to the width by a factor of = | >~1.41. | | Yes. Proportional to the width. Not to the square of the width. | | >Or, the width/thickness is always 70.7106781186547524400844362105198% of | >the bevel/miter. | | 70.7 % is plenty close enough. | | -- | Regards, | Doug Miller (alphageek at milmac dot com) | | Nobody ever left footprints in the sands of time by sitting on his butt. | And who wants to leave buttprints in the sands of time?
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You missed the point rather dramatically, I'm afraid. You wrote that the width of the miter was proportional to "the square of the width of the board".
This is false.
It is proportional to the *width* of the board. Period. Not the square of its width.
You then compounded this error by repeating it with respect to thickness, and bevels.
And now you've compounded it still further by showing that, in addition to your difficulties with mathematics, you also have some reading comprehension issues.
-- Regards, Doug Miller (alphageek at milmac dot com)
Nobody ever left footprints in the sands of time by sitting on his butt. And who wants to leave buttprints in the sands of time?
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If you mean a miter cut, the length of the miter is
the root of two times the square of the width of the board.
If you mean a bevel cut, the length of the bevel is
the root of two times the square of the thickness of the board.
1 inch wide = 1.4142135623730950488016887242097 2 inch wide = 2.8284271247461900976033774484194 3 inch wide = 4.24264068711928514640506617262909 4 inch wide = 5.65685424949238019520675489683879
It appears the bevel/miter is proportional to the width by a factor of ~1.41. Or, the width/thickness is always 70.7106781186547524400844362105198% of the bevel/miter. _________________________________________________________
Dougie, you said
| You missed the point rather dramatically, I'm afraid. You wrote that the width | of the miter was proportional to "the square of the width of the board".
I don't think so. No where in the preceding, which I include herewith for clarity, did I state what you saw.
Better get your eyes checked. Your geekiness leaves much to be desired. You might, however, be in line for the "Conehead" awards. ________________________________________________________ --
PDQ --
| >Picky, picky, picky. | > | >If you want to play those games, Doug: | > | >"Bevel" is described as "the angle formed at the juncture of two non = | >perpendicular surfaces." | > | >"Miter" could mean "a tall ornamental liturgical headdress" worn by some = | >members of the clergy, or it could mean, as it does in this case, = | >"either of the surfaces that come together in a miter joint". | > | >If you want to play with polygonal surfaces, why not say so? "board = | >_not_ squared" is so imprecise. | > | >I guess your problem must lie with your inability to visualize the = | >position of the board within its frame of reference. | | You missed the point rather dramatically, I'm afraid. You wrote that the width | of the miter was proportional to "the square of the width of the board". | | This is false. | | It is proportional to the *width* of the board. Period. Not the square of its | width. | | You then compounded this error by repeating it with respect to thickness, and | bevels. | | And now you've compounded it still further by showing that, in addition to | your difficulties with mathematics, you also have some reading comprehension | issues. | | -- | Regards, | Doug Miller (alphageek at milmac dot com) | | Nobody ever left footprints in the sands of time by sitting on his butt. | And who wants to leave buttprints in the sands of time?
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{ *VIEW IN A FIXED-PITCH FONT* e.g. 'fixedsys' on a Windows PC ]

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

Actually, you *did*. And you even quoted those _exact_words_, above. "For clarity", the occurrences of the indicated words have been marked, so that the vision-impaired can locate them.

"Speak for yourself, John" would seem to apply.

You're the leading candidate for the pseudo-"Ronald McDonald" award. (The one named for the _original_ 'big red hair' circus entertainer, made Famous by Larry Harmon.)
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Guess you never pretended to be logical.
I said root(2(width*width)).
My professors told me that, in the parlance, root equates to square root. It is just a convenient form thereof.
Assuming you can comprehend the above, your underscore, via a caret, is the same. I only wish I had a proper symbol on this pig.
--
PDQ --
| >If you mean a miter cut, the length of the miter is | > | >the root of two times the square of the width of the board. | ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ | ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ | > | >If you mean a bevel cut, the length of the bevel is | > | >the root of two times the square of the thickness of the board. | ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ | ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ | > | >1 inch wide = 1.4142135623730950488016887242097 | >2 inch wide = 2.8284271247461900976033774484194 | >3 inch wide = 4.24264068711928514640506617262909 | >4 inch wide = 5.65685424949238019520675489683879 | > | >It appears the bevel/miter is proportional to the width by a factor of ~1.41. | >Or, the width/thickness is always 70.7106781186547524400844362105198% of | >the bevel/miter. | >_________________________________________________________ | > | >Dougie, you said | > | >| You missed the point rather dramatically, I'm afraid. You wrote that | >the width | >| of the miter was proportional to "the square of the width of the board". | > | >I don't think so. No where in the preceding, which I include herewith | >for clarity, did I state what you saw. | | Actually, you *did*. And you even quoted those _exact_words_, above. | "For clarity", the occurrences of the indicated words have been marked, | so that the vision-impaired can locate them. | | | >Better get your eyes checked. Your geekiness leaves much to be desired. | | "Speak for yourself, John" would seem to apply. | | >You might, however, be in line for the "Conehead" awards. | | You're the leading candidate for the pseudo-"Ronald McDonald" award. | (The one named for the _original_ 'big red hair' circus entertainer, made | Famous by Larry Harmon.) |
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Bzzzt! Thank you for playing.
That may have been what you _intended_ to say (I'll not speculate on *that*), but it is *not* what you actually wrote. You wrote the English words for "root(2) * width*width"
"root" is a 'higher priority' "operator" than 'times', and the associativity is left-to-right.
Given that what you wrote above is what you actually intended to say originally, you omitted a critical phrase from your scrivening. The words "the quantity" was required after 'root of"

No argument on _that_ point.
Did your professors bother to teach you about "reduction" to simplest form?
Did your professors not teach you how *stupid* it is to do two multiplies and a (calculated) square-root when the exact same result can be obtained via a single multiply of a constant

Tell me, just how would you express _in_words_, "root(2) * (width*width)" then?

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This is more fun that actually applying myself to wood.
Have you never given any thought to the order of qualification inherent in the utilization of "of"?
The resultant of any number multiplied by itself is the square of that number.
ergo: miter length = root (two(thickness squared)) .
Amazing what is lost as a result of the "whole language" system.
--
PDQ --
| >| >If you mean a miter cut, the length of the miter is | >| > | >| >the root of two times the square of the width of the board. | >| ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ | >| ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ | >| > | >| >If you mean a bevel cut, the length of the bevel is | >| > | >| >the root of two times the square of the thickness of the board. | >| ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ | >| ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ | >| > | >| >1 inch wide = 1.4142135623730950488016887242097 | >| >2 inch wide = 2.8284271247461900976033774484194 | >| >3 inch wide = 4.24264068711928514640506617262909 | >| >4 inch wide = 5.65685424949238019520675489683879 | >| > | >| >It appears the bevel/miter is proportional to the width by a factor of ~1.41. | >| >Or, the width/thickness is always 70.7106781186547524400844362105198% of | >| >the bevel/miter. | >| >_________________________________________________________ | >| > | >| >Dougie, you said | >| > | >| >| You missed the point rather dramatically, I'm afraid. You wrote that | >| >the width | >| >| of the miter was proportional to "the square of the width of the board". | >| > | >| >I don't think so. No where in the preceding, which I include herewith | >| >for clarity, did I state what you saw. | >| | >| Actually, you *did*. And you even quoted those _exact_words_, above. | >| "For clarity", the occurrences of the indicated words have been marked, | >| so that the vision-impaired can locate them. | >| | >| | >| >Better get your eyes checked. Your geekiness leaves much to be desired. | >| | >| "Speak for yourself, John" would seem to apply. | >| | >| >You might, however, be in line for the "Conehead" awards. | >| | >| You're the leading candidate for the pseudo-"Ronald McDonald" award. | >| (The one named for the _original_ 'big red hair' circus entertainer, made | >| Famous by Larry Harmon.) | >| | |
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PDQ, to be consistent with your first post, you should use: ergo: miter length = root (two(width squared)) OR ergo: bevel length = root (two(thickness squared))
Sorry, given how the thread was going I couldn't help myself.
BadgerDog
This is more fun that actually applying myself to wood.
Have you never given any thought to the order of qualification inherent in the utilization of "of"?
The resultant of any number multiplied by itself is the square of that number.
ergo: miter length = root (two(thickness squared)) .
Amazing what is lost as a result of the "whole language" system.
--

PDQ
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| PDQ, to be consistent with your first post, you should use: | ergo: miter length = root (two(width squared)) | OR | ergo: bevel length = root (two(thickness squared)) | | Sorry, given how the thread was going I couldn't help myself. | | BadgerDog | | Some days one just can't seem to do more than survive. By the time I got to this point, I almost didn't even care how the board was positioned.
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PDQ says...

You should find another job other than trying to prove you are smarter than everybody else. It's a crowded field. Besides, if you are going to try to sound like Einstein, you should at least be right.
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| PDQ says... | | > This is more fun that actually applying myself to wood. | > | > Have you never given any thought to the order of qualification inherent in the utilization of "of"? | > | > The resultant of any number multiplied by itself is the square of that number. | > | > ergo: miter length = root (two(thickness squared)) . | > | > Amazing what is lost as a result of the "whole language" system. | | You should find another job other than trying to prove you are smarter | than everybody else. It's a crowded field. Besides, if you are going | to try to sound like Einstein, you should at least be right.
Can't say as I was/am trying to prove myself "smarter than the average bear".
I was just replying to a couple of pedants.
--
PDQ --
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And you still don't realize where you went wrong.
-- Regards, Doug Miller (alphageek at milmac dot com)
Nobody ever left footprints in the sands of time by sitting on his butt. And who wants to leave buttprints in the sands of time?
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Geometry is proveable. I took a course where we began with Peano's postulates and from that derrived all of elementary Calculus--that stuff with differentials and integrals you might have hit in college.
It took two semesters and was BRUTAL.
Lemme see if I remember . . .
There is a number Zero
(MANY things snipped)
. . . which proves that the limit exists about a point x0.
Easy peasy.
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Analytical geometry/spherical trig: I think that was the name of the only college course I couldn't pass... (I did great in algebra, though!)
Dave
Charles Krug wrote:

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