Figurin' angles......

[Dennis Peacock]

OK.....Call me stupid and slap me silly........but how do you figure out what an angle is?

Let's say that I have a "shape" that I want to inlay. I need to figure out what an angle is so I can cut it at least "close" the first time. If I know the length of one side of the triangle and from the same common point I know the length of the other side and I can measure and get the length of the 3rd side......What is the simple method if figuring out what the angle is between say A and B if I know what the lengths are for A, B & C.


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So what is the "shop" formula for figuring out the angle.? Now, I flunked Algebra in highschool...so take it easy......

[Terry Hatfield]

d,

Beats me. :D

Is that CD cabinet kicking your butt or what??? :eek:

t

[Ed Falis]

Dennis,

I'm assuming you have a trig calculator available (the windows Calculator program that comes with the OS can do it if you set it to scientific mode).

The general formula, laid out so that you want to find angle C, give sides a, b and c (where side c is opposite angle C) is:

C = arccos ((a^2 + b^2 - c^2) / (2 x a x b))

a^2 mean "a squared"
arccos means the inverse cosine

To test whether you've got it, try it with a=3, b=4, c=5. You should get 90 degrees as the result (make sure the calculator is set to use degrees instead of radians).

Feel free to email if you need to.

- Ed

[Dennis Peacock]

d,

Beats me. :D

Is that CD cabinet kicking your butt or what??? :eek:

t

<B>Yup!!!!</b> Figurin' angles and lengths and "stuff" is really pushing me WAY beyond my abilities. I guess some folks call that "being stretched". Great to see you posting again.!!!!!

Back to the shop for more torture of angles. I now have to try Eds formula out.

<B>THANKS ED...I'll give'er a try..!!!</b>

[Keith Outten]

Dennis,

If you know the length of two sides of a triangle the formula is;

a squared plus b squared = c squared (long side)

[John Wadsworth]

That formula (the famous Pythagorean theorem proved it) is for right triangles only--that is, the angle between the a and b sides has to be 90 degrees--and that's the angle Dennis is trying to solve for!

[Ed Falis]

Dennis,

Like I said, give a yell if you need to. It's not as tough as it looks once you've done it once or twice.

- Ed

ps I failed hs algebra the first time, and wound up with a math degree 10 years later. Was just a matter of deciding I actually had a use for the math. I suspect you may be the same kind of guy. ;-)

[Dennis Peacock]

That formula (the famous Pythagorean theorem proved it) is for right triangles only--that is, the angle between the a and b sides has to be 90 degrees--and that's the angle Dennis is trying to solve for!

John,

Here is what I am trying to figure out:

Side A = 2.92"
Side B = 2.92"
Side C = 3"
Side D = 2.92"
Side E = 2.92"
(Diamond Shape (3" wide by 6" tall))

What is the miter angle so that side A and B will mate (point of the diamond)?
What is the miter angle so that side A will mate with Side D (one side of the diamond)?
etc.....for Side B to Side D.

I know someone here can figure that out....BUT....I want to know how to simply figure stuff like that out without having to post a question here to find the answer. This "should" be some kind of "shop math" that I just can't seem to get the hang of. Large Scale computer systems I can do, Home PC systems I can do.....but this angle figurin' has got the best of me. I know if I can just see the equation and see it done just once, I am sure I will take it to the shop to use from here on out and will be a better ww'er because of learning how to figure on my own.

[Dennis Peacock]

Dennis,

Like I said, give a yell if you need to. It's not as tough as it looks once you've done it once or twice.

- Ed

ps I failed hs algebra the first time, and wound up with a math degree 10 years later. Was just a matter of deciding I actually had a use for the math. I suspect you may be the same kind of guy. ;-)

Ed,

You are exactly correct!!!!!! I "want" to learn how to do this on my own since I am trying to learn how to do things better instead of spending 3 days cutting pieces of wood trying to find the correct angle.

[Ed Falis]

John,

Here is what I am trying to figure out:

Side A = 2.92"
Side B = 2.92"
Side C = 3"
Side D = 2.92"
Side E = 2.92"
(Diamond Shape (3" wide by 6" tall))

What is the miter angle so that side A and B will mate (point of the diamond)?
What is the miter angle so that side A will mate with Side D (one side of the diamond)?
etc.....for Side B to Side D..

OK, Boss, let's take a step-by-step shot at it.

a= b = 2.92
c = 3

C = arccos ((a^2 + b^2 - c^2) / (2 x a x b))
= arccos (( 2.92 ^2 + 2.92 ^2 - 3^2) / 2 x 2.92 x 2.92))
= arccos ((2 * 2.92^2 - 3^2) / 2 * 2.92 ^2)
= arccos ((2* 8.536 - 9) / 2*8.526)
= arccos (8.053/17.053)
= arccos (.47227....)
= 61.82 degrees

and I'm guessing 62 degrees will do the job.

- Ed

[Dennis Peacock]

OK, Boss, let's take a step-by-step shot at it.

a= b = 2.92
c = 3

C = arccos ((a^2 + b^2 - c^2) / (2 x a x b))
= arccos (( 2.92 ^2 + 2.92 ^2 - 3^2) / 2 x 2.92 x 2.92))
= arccos ((2 * 2.92^2 - 3^2) / 2 * 2.92 ^2)
= arccos ((2* 8.536 - 9) / 2*8.526)
= arccos (8.053/17.053)
= arccos (.47227....)
= 61.82 degrees

and I'm guessing 62 degrees will do the job.

- Ed

<B><font size=+2>You-Da-Man Ed!!!!</font></b>

I just did what you did on the Winders Calcu-muh-lator and I got your answer!!! I'm tickled..!!!!! The LOML says that I will go around figuring up EVERYTHING around the house now..!!!! :D

Many thanks to ALL of you for your assistance, advise and knowledge!! I'm a happy ww'er now......Well.......gotta run and start figurin' out every angle in the shop.......now let's see.......If the rafters are a length of.........

[Ed Falis]

Math can be a powerful thing for good in the right hands, Dennis. Go get 'em.

- Ed

[Phil Phelps]

I slap your punkin' head around to a 60 degree angle for not showin' up at Jenkins' :D

[Dennis Peacock]

I slap your punkin' head around to a 60 degree angle for not showin' up at Jenkins' :D

Sorry Phil......

I was gonna come to Jenkins' BBQ but I had to work at "work" for a while today. I will be there next year...even if I have to taken vacation time to be there. I just couldn't get off this time to make the trip. I hope all of you had a grand time!!! Post some pics....or did you forget your camera? ;)