View Full Version : Math Question

Matt Ocel

10-21-2008, 6:28 PM

I had to build a angled platform today. Because i'm a carpenter, sometimes my feeble little brain cause me to take the practical approach (more time)instead of the mathamatical approach (less time).

Maybe someone out there can help.

The platform is 36" wide.

The angle is 26 degrees.

On the inside part of the angle, it is 12" from square to short point. Then 12" from short point to square.

Now, my practical approach is to snap this out on a sheet of ply To find the lengths of the outer side. This works great. But I need to broden my horizons.

What is the mathamatical equation for this?

Tom Veatch

10-21-2008, 6:43 PM

If I'm understanding you correctly, you're asking for the relationship between the hypotenuse (angled side) and the legs (horizontal and vertical sides) of a right triangle.

If that's correct, the relationship is: The square of the hypotenuse is equal to the sum of the squares of the two sides.

In this case, taking the 36" dimension to be the horizontal side and 12" to be the vertical side, the hypotenuse (outer side?) would be 37.947" (appx 37 15/16")

But, that assumption could be wrong because a right triangle with sides of 12" and 36" has an acute angle of ~18.5 degrees, not 26. So obviously I'm not reading your question correctly.

BTW, there's absolutely nothing wrong with the "practical" approach. Many times it's a lot quicker, and sufficiently accurate, to lay it out as you described. If you need precision down to the thousandth of an inch, the math approach might be more accurate, but that's rarely needed in day-to-day woodworking.

Jason Beam

10-21-2008, 6:45 PM

Not quite sure I follow what you're after ...

But ...

It sounds a little like yer after the Hypotenuse of a right triangle. That is, you have a right angle and one side goes up 12" and the other side goes out 36" and you want to know the length of the line that connects the ends and completes the triangle. That's the Hypotenuse.

To find it, you square both sides, add them, then take the square root of that sum:

12*12 + 36*36 = 1440 inches.

The sqrt of that is: dang near 38 inches - 37.9473319 (linky (http://www.google.com/search?hl=en&q=square+root+of+1440&btnG=Google+Search&aq=f&oq=))

If that's not what you're after, I apologize :)

Jason Beam

10-21-2008, 6:46 PM

But, that assumption could be wrong because a right triangle with sides of 12" and 36" has an acute angle of ~18.5 degrees, not 26. So obviously I'm not reading your question correctly.

Aha! I hadn't even thought to check that. I guess we're both off!

Scott Erwin

10-21-2008, 8:23 PM

If it is for a right angle. Then the prior posters are correct.

a*a + b*b = c*c (a squared plus b squared equals c squared)

This is called the Pythagorean Theorem.

http://mathworld.wolfram.com/PythagoreanTheorem.html

Not a right angle....then it is called an Oblique Triangle. But there is no formula I am aware of that will get you the right answer...but there may be some propeller heads that will prove me wrong...

Matt Ocel

10-21-2008, 8:25 PM

This should explain it better.

Doug Shepard

10-21-2008, 8:26 PM

When all else fails, draw it in Sketchup, rotate it to the desired angle and use the dimension tool to measure it.

Matt Ocel

10-21-2008, 8:43 PM

When all else fails, draw it in Sketchup, rotate it to the desired angle and use the dimension tool to measure it.

That would fall under the "not" practical approach on the job site.

Ben Davis

10-21-2008, 8:47 PM

i mis-typed!

just shy of 20 5/16". 20.311"

I'll post a picture in a sec.

Matt Ocel

10-21-2008, 8:53 PM

just shy of 44 5/16". 44.311"

I'll post a picture in a sec.

Can't be.

and to answer your question, it is I.

Peter Quadarella

10-21-2008, 8:58 PM

Matt, the easiest way to do this is to subdivide the shape into two rectangles and two triangles (one rectangle and one triangle on each mirror image side of your shape).

Then you will know that the ? side is 12" plus one side of a right triangle. To get the missing side of the triangle you would:

1. You already have one side of the triangle - 36" To get the other side, you would need to figure out the small angle on the triangle (call the angle X). Then take Tan (X) [tangent of x], this is a mathematical function that you can look up. This will give you a decimal.

Tangent X = opposite side /adjacent side. Once you have the adjacent side (36"), and the value of Tan (X), you can solve for the opposite side.

Then you can add that side of the triangle to the side of the rectangle and you have your "?". Unfortunately, I cannot solve Tan (X) for you because the angle in your picture is wrong (it has to be something in between 90 degrees and 180 degrees from looking at the picture).

It is really hard to explain geometry/trigonometry without being able to draw, sorry. I am also working off of 21 year old memories so forgive me.

Matt Ocel

10-21-2008, 9:08 PM

Peter -

If it helps, and remember this is a carpenter talking, the way I came up with 26 degrees, is by using my speed square, a couple straight edges, and so on, and so on.

Ultimately I cut my mitres each at 13 degrees.

26 off of 180, maybe it is a 156 degree angle????

Once again, a nail bender talking.:D

Doug Shepard

10-21-2008, 9:10 PM

What Peter said. You either need to know the angle of each of the triangles marked A & B or you need the length of the hypotenuse shared by those. Then it's simple trig functions to get the height of the portion along the edge and add to 12".

99115

Ben Davis

10-21-2008, 9:20 PM

Okay. I got my pic posted above with all of the math. See Post #9.

Tom Veatch

10-21-2008, 9:22 PM

99115

? = 12 + 36tan(13) = 20.31125488052027202293324208508

20 5/16"

Matt Ocel

10-21-2008, 9:24 PM

Ben, I think your onto something.

Now if you can speak a little slower, I know I can get my feeble little brain around this.

If you know what I mean.;)

Matt Ocel

10-21-2008, 9:33 PM

OK - I think I got it.

The missing piece is how does X=(36")(tan13 degrees) = 8.311 ???????

Doug Shepard

10-21-2008, 9:34 PM

OK. The pic was throwing me as it didn't look like the same angle on both pieces. Should have read the text closer. 13 degrees on each so the 20.3 should be good.

Peter Quadarella

10-21-2008, 11:46 PM

Thanks for the drawing Doug. Since the line between A and B, if extended, bisects the angle X, we know then that the angle we are looking for, Y = 90 - (1/2)X.

If X were 156 then Y = 12. Tan (12 degrees) * 36 = 7.65". So the ? would be 12 + 7.65 = 19.65".

I'm not 100% clear if X really is 156 degrees, so if it's different the answer would change obviously.

I'm not clear on how Ben got his angle measurements, so if he knows something I don't then use his measurements. Somehow he knows alpha is 26 degrees, which if true all his logic follows.

Peter Quadarella

10-21-2008, 11:47 PM

OK - I think I got it.

The missing piece is how does X=(36")(tan13 degrees) = 8.311 ???????

Matt, for stuff like this, if you don't have a good calculator handy just type this into a google search:

(tan(13 degrees))* 36

Doug Shepard

10-22-2008, 5:36 AM

...

I'm not clear on how Ben got his angle measurements, so if he knows something I don't then use his measurements. Somehow he knows alpha is 26 degrees, which if true all his logic follows.

I think ultimately Matt's statement straightens this out. The 13 deg isn't really apparent from the pic.

...

Ultimately I cut my mitres each at 13 degrees.

26 off of 180, maybe it is a 156 degree angle????

...

Peter Quadarella

10-22-2008, 9:00 AM

Ah thanks, and thanks to Matt for getting me to clear some cobwebs from my head :).

Ben Davis

10-22-2008, 12:38 PM

I think ultimately Matt's statement straightens this out. The 13 deg isn't really apparent from the pic.

The 13* angle is 1/2 of the stated 26* rotation. The top rectangle has been rotated CCW about it's left lower point by 26*. Bisecting this angle (alpha) gives you the 13*. I'm used to using variables for everything until the very last step. Then I plug in the actual numbers. It makes simplication of complex equations easier.

Let's make it interesting and make it a curvy shape! Now we're cooking with gas!

Ben Davis

10-22-2008, 12:44 PM

OK - I think I got it.

The missing piece is how does X=(36")(tan13 degrees) = 8.311 ???????

You're right on the money Matt. The tangent of an angle alpha in a right triagle is the length of the side opposite the angle divided by the length of the side adjacent to the angle.

I guess some terminology is needed. Three sides to a right triangle. Opposite. Adjacent. Hypotenuse.

Sin (alpha) = Opp/Hyp

Cos (alpha) = Adj/Hyp

Tan (alpha) = Opp/Adj

From there you can solve for the angle or the length of a side. As an aside, all of this makes much much much more sense when working in Rads (radians) vs. degrees. The math and numbers are expressed in fractions or multiples of Pi. Things get nice and simple that way. Unfortunately, it's essentially useless in real life measurements. You would never say, "The length of that side is 4-Pi feet long."

There are other really quite interesting things that happen with these trig functions as we play some simple algebra with them. For example, evaluate sin(x)/x. Pick values just greater than and just less than 0. Basically with some other math rules, you can prove that sin(0)/0 = 1, thus allowing you to now divide by zero. Weird huh?

Chris Kennedy

10-22-2008, 12:58 PM

If it is for a right angle. Then the prior posters are correct.

a*a + b*b = c*c (a squared plus b squared equals c squared)

This is called the Pythagorean Theorem.

http://mathworld.wolfram.com/PythagoreanTheorem.html

Not a right angle....then it is called an Oblique Triangle. But there is no formula I am aware of that will get you the right answer...but there may be some propeller heads that will prove me wrong...

If it isn't a right triangle, you can use the law of cosines:

c^2 = a^2 + b^2 - 2ab cos C

where C is the angle opposite side c (equivalently, the angle between a and b). If C is 90 degrees, then the cosine is zero and you get good 'ol Pythagoras.

In essence, the law of cosines gives you a "correction" factor when the triangle isn't a right triangle.

Cheers,

Chris

Chris Padilla

10-22-2008, 7:35 PM

There are other really quite interesting things that happen with these trig functions as we play some simple algebra with them. For example, evaluate sin(x)/x. Pick values just greater than and just less than 0. Basically with some other math rules, you can prove that sin(0)/0 = 1, thus allowing you to now divide by zero. Weird huh?

sin(x)/x is the sinc function. It also obeys L'Hopital's Rule where you can differentiate the upper and lower function independent of each other when doing limits....

Ah, I guess I did learn something in calculus in college.... :D

Matt Ocel

10-22-2008, 8:16 PM

Whoa!

sin? tan? ^? alpha?

Looks like l'll be buying more chalk for my chalk line.:D

I knew I should'nt have skipped school the day they were teachin math.

Peter Quadarella

10-22-2008, 8:21 PM

L'Hopital's Rule! Wow, never expected to clear that cobweb on a woodworking forum :D.

Doug Shepard

10-22-2008, 8:46 PM

L'Hopital's Rule! Wow, never expected to clear that cobweb on a woodworking forum :D.

No foolin. Can't even remember how many girls I picked up using the old L'Hopital's rule.:D

That one's just as dusty as Stokes theorem, Green's theorem, Euler's theorem, Rolles theorem and all the other math I've forgotten.

Ben Davis

10-22-2008, 8:52 PM

Whoa!

sin? tan? ^? alpha?

Looks like l'll be buying more chalk for my chalk line.:D

I knew I should'nt have skipped school the day they were teachin math.

It's not that bad after you see the pictures and then see how someone uses the rules correctly. The problem in schools is that they make it hard. It's just a rule.

Anyone want to talk about line integrals and gradients? Now we're REALLY clearing out the cobwebs!

-Ben

Chris Padilla

10-23-2008, 3:44 PM

Hey, Ben started it!!! The only reason I remember L'Hopital's rule is that a buddy in grad school use to like to say: "Is it L'Hopitalic?" That stuck and I've never forgotten the rule! :D

Ah, Green's Functions. My wife's whole dissertation was one large dyadic Green's Function.

How about integration by parts? Maxwell and Poisson's equations? Hyperbolic parabaloid anyone? Leibiniz's Rule? Lagrange multipliers? Laplace?

:D :D :D

Chris Kennedy

10-24-2008, 11:48 AM

I'll see your integration by parts and raise you a trig sub:D.

Power series? Fourier series? Or my favorite, Lie algebras (pronounced just like Lie Nielsen).

Cheers,

Chris

Peter Quadarella

10-24-2008, 1:01 PM

I got nothing. It's all wisps of smoke now :confused:. You guys have excellent memories!

Doug Shepard

10-24-2008, 1:17 PM

I... Or my favorite, Lie algebras (pronounced just like Lie Nielsen).

Cheers,

Chris

How appropriate that you bring that horror up as Halloween approaches.:eek:

That got covered in a truly mindbending course called Abstract Algebra along with other scary monsters. Nothing but writing proofs proving groups, fields, etc. I still have no idea how I managed to pass that course with an A, cuz for the life of me I couldn't tell you even then what it was all about.

Chris Kennedy

10-24-2008, 5:33 PM

Doug:

I have to ask. Where did you run into Lie algebras in an Abstract class? Usually they are reserved for graduate classes. Do you remember your prof's name?

(I am an algebraist and specialize in a particular type of Lie algebra -- that's why I am asking.)

Cheers,

Chris

Matt Ocel

10-24-2008, 5:40 PM

Lie algebras is far overated

I Personally prefer THE CAMPBELL BAKER HAUSDORFF FORMULA.:D

Doug Shepard

10-24-2008, 6:14 PM

Chris

I was finishing up on a Math/CS degree at http://www.ltu.edu/. Cant recall the profs name. This was about 12 years ago. I looked on their online faculty directory and while some of the same ones I had for other classes are still there, nothing rings a bell for that Abstract course so he may have retired. I'm sure we were probably just given a sampler problem or two. That whole course seemed like it jumped around through numerous topics that didn't seem all that related. It's strange that of all my math courses, it had the thinnest textbook yet was probably the hardest course. I think around 30 people started the course and it ended up around 8-10. Most of us had to pick up a few other books just to get through it. I just recall about 2 weeks of the course spent on a couple Lie Algebra problems that were excruciating. Thankfully I dont think any of it was on the exams that I recall. I'd rather vacation in sunny Baghdad than repeat that course.

Ben Davis

10-24-2008, 8:06 PM

Okay. You guys went high and right from where I jumped off the math band wagon. I do remember taking a number theory class in college, but I never understood a single thing in there (no kidding.. not one). I was much more of an applied math... al la engineering. Maxwell equations, Lagrangians, etc.

I always wanted to know more about the groups, sets, rings, et al, but everything I tried to read was gibberish to me. Without someone explaining this stuff, I'm lost. C'est la vie. No hope now to master it.

"There is safety in numbers."

Chris Kennedy

10-25-2008, 3:23 PM

Lie algebras is far overated

I Personally prefer THE CAMPBELL BAKER HAUSDORFF FORMULA.:D

Oh -- gonna pull CBH on me? I may just have to unleash a Killing form.:D

Cheers,

Chris

Chris Kennedy

10-25-2008, 3:26 PM

Doug,

I can think of only three people in the world who would be sadistic enough to unleash Lie algebras on undergraduates -- one is my advisor from U of M. Another is one of his students, whose name I cannot remember, but I know I would recognize his name if I read it. I think he teaches at Eastern Michigan, though.

Cheers,

Chris

Doug Shepard

10-25-2008, 3:44 PM

Could be the evil twin of your advisor. While I recall almost nothing about the math of it (even then) I earned his scorn over the name. He had a real peeve whenever anyone mispronounced it and would sternly correct it. About midway through that little 2 week detour into Hell, he announced that a few more people had dropped the course. I quipped back: No Lie?

Seemed to get a chuckle from the rest of his victims but he wasn't laughing.

Matt Bickford

10-27-2008, 2:32 PM

Gor Blimey, Nerd alert!

Steven DeMars

10-27-2008, 6:34 PM

I am math challenged unless there is a $ sign or a % in it . . .

That is why I use AutoCAD for everything . . .

Steve

Karl Brogger

10-27-2008, 8:34 PM

Matt- Just out of curiosity, what particular "tavern" were you solving the worlds problems in?

I had to open Word to spell curiosity BTW, I get dumber every day.

Matt Ocel

10-28-2008, 4:28 PM

Matt- Just out of curiosity, what particular "tavern" were you solving the worlds problems in?

I had to open Word to spell curiosity BTW, I get dumber every day.

"Babes"

In Lakeville

Karl Brogger

10-28-2008, 6:53 PM

"Babes"

In Lakeville

LOL, yep. Been three sheets to the wind there a few times. Used to work not too far from there at Designed Cabinets. If you're ever desperate for work, that should be last place you turn in your application at. You're better off checking in at a rendering plant, waste treatment facility, you get the idea.

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