finding angles (math question)

[kshellrazor]

Please view the attached. Is there a formular for determining the angles where I have marked in red? Say for ****s and giggles I was using 1 inch square tubing and needed to cut the angles I have highlighted in red. Put in any dimensions for the sides. Does not matter. I just need to know if there is someway I can figure them accurately using algebra, trig, basic math, rocket science, anything!!! Thanks!!!

[yhprum]

Easist way is to google trig calculator, pick one and plug in the numbers you know.

[kshellrazor]

Easist way is to google trig calculator, pick one and plug in the numbers you know.

That is like finding a needle in a haystack! - most of them won't account for something in 2 or 3 dimensions, like tubing.

I have a feeling somebody in here does this all the time.

[kshellrazor]

Do people clamp the tubing, and mark the angle. This is a cheap way, but I was thinking there might be a more 'professional' way. ha -

....especially in close quarters where using some sort of template system is not possible.

[Tailshaft56]

If you know the length of the sides trig functions will allow you to compute the angles. It's been too many years for me to remember but a review of most shop math books should spell it pretty quickly. If you want to compute the lenth of the hypotinous pythagorem therem will do it. a squared plus b squared = c squared where a and b are the sides of a right triangle.

[riley mcmillan]

side opposite divided by side adjacent gives the tangent of the angle you are looking for. This works for either angle. Just remember that the long side is the hypotenuse. Then you can either look it up in trig function chart or use a calculator with trig functions.

[zapster]

There is a math forum on this site :confused:

...zap!

[bluestreak]

I don't have any math answers, but i made a little tool I use all the time when I have to fab stuff up on the jobsite. It's just two pieces of 1"x1/4" flat bar (you can use smaller stock if you like) with a 3/8 hole drilled through both pieces about 1/2" from the end. (You can make these as long as you need. Mine are about a foot.) Secure the pieces together with a short bolt and a wing nut. Round the corners closest to the bolt with a grinder. Set the pieces on edge at the place you need to find your angle and open them like scissors until you get the desired angle. Tighten the wingnut. Now you can transfer your angle exactly and directly to the piece you need to fit. If you need to record the specific degree, you can set the "angle finder thingy" next to a protractor or a speed square.

[Weldordie]

I have found that a cheap angle finder from Ace Hardware keeps me from having to do any mathematical figuring of angles.

(Apparently, Bluestreak and I posted at about the same time. He has, essentially, described the tool that I purchased at Ace.)

[WillieB]

you can use a protractor but it's not as accurate as trig. Tables of trigonometric functions are easily available so remember, sine = O/H, cosine = A/H and tangent = O/A where O is the side opposite the angle, A is the side adjacent to the angle and H is the hypotenuse. Do the division and look on the trig table for the sine, cosine or tangent which matches the angle you're looking for.

[smithboy]

http://argyll.epsb.ca/jreed/math9/strand3/3103.htm
Here is a cool little slide dohicky that lets you find the angles of any size triangle with interger values for the legs. See the middle of the page. Lots of other trig calculators also...

[bluestreak]

About the "angle thingy"....I forgot to tell you to tack the bolt head in place so it doesn't spin. And those other fellows are correct, trig will get you a much better answer. I just don't normally keep a calculator in my bucket. It might get smashed by a tool and I would just end up setting a book on fire.

[tapwelder]

I guess we are assuming that the unmarked angle is 90 degrees? If so then the basic trig functions and calculator or chart would work.

I have no problem clamping, marking then cutting whenever possible. Next method would be a bevel square to figure the angle. Finally, mathematically. Many times I calculate mathematically, just to check.

[kshellrazor]

Man, love this forum. Thanks to all. I got some sites to look at and an old calclator to dig up. Much thanks!

[SSBN727]

Does anybody remember that "old indian" named SOH CAH TOA???:eek: :eek: ;)

Respectfully
SSBN727

[lotechman]

Every tool box should have a calculator in it. It's good for checking your pay also. If one is weak in trig and math then just remember rise over run. That fraction is the tangent value and is a ratio. You can use that ratio and multiply by twelve to find out what the bevel angle is in twelve inches. Using a two foot framing square and making your angle by bevels in twelve is very accurate.
Soon you wil be playing with compound angles DANGIT! From hints from others I hope to do a solution thid weekend for the compound proplem I posted. It is similar to this problem only multi step.

[Flywheel]

Please view the attached. Is there a formular for determining the angles where I have marked in red? Say for ****s and giggles I was using 1 inch square tubing and needed to cut the angles I have highlighted in red. Put in any dimensions for the sides. Does not matter. I just need to know if there is someway I can figure them accurately using algebra, trig, basic math, rocket science, anything!!! Thanks!!!

Measure the long edge and short edge, then mark the peice to cut, draw a line and cut.....

[hdwood]

Every tool box should have a calculator in it. It's good for checking your pay also. If one is weak in trig and math then just remember rise over run. That fraction is the tangent value and is a ratio. You can use that ratio and multiply by twelve to find out what the bevel angle is in twelve inches. Using a two foot framing square and making your angle by bevels in twelve is very accurate.
Soon you wil be playing with compound angles DANGIT! From hints from others I hope to do a solution thid weekend for the compound proplem I posted. It is similar to this problem only multi step.

If you have a framing square you will not need any math. This tool was designed to cut the angles required to frame a roof. The rise and run will relate directly if the dimensions are less than the length of your frameing square. If not here is a site for more details. http://www.homefocus.com/410/framing_square_know-how.htm

[backuproller]

i see alot of questions about pipe and tubing. a pipefitters blue book has all the equations you need. if you look in the pawn shops or talk to the maint. man in the plant, they will probally have one laying around.
it will cover angles, off-sets, rolling off-sets. anything you need to work with pipe and tube.

[Weldtek]

All,

I am new here. I know this is a little late, but. I used a little system in skool yeers ago to remember the basic trig functions. The way I got it was " Some Crazy Thing - Oscar Has A Hairy Old A$$.

Sin = Opposite over Hypotenuse
Cos = Adjacent over Hypotenuse
Tan = Opposite over Adjacent

Also, I would lay out the lengths from the centerline of the tube, scribe the angles and cut the mitres.


Weldtek
There's no such thing as common sense!

[fwalz3]

I have found that a cheap angle finder from Ace Hardware keeps me from having to do any mathematical figuring of angles.


When I first read this I thought it said "cheap angle grinder" which I guess, if all else fails, will work too!!! :laugh:

[Weldordie]

When I first read this I thought it said "cheap angle grinder" which I guess, if all else fails, will work too!!! :laugh:

The times that I can't find my cheap angle finder, I resort to using my cheap angle grinder. You are right... it does work!:cool:

[clive]

Does anybody remember that "old indian" named SOH CAH TOA???:eek: :eek: ;)

Respectfully
SSBN727
Yep, I learned about him when I was an apprentice many moons ago, I still use it to teach trig.
It's a magic word actually, you utter it and you see a whole class just go blank, amazing effect.

[steve45]

Angle 2 is arctan(vertical/horizontal) or, tan-1 (vertical/horizontal) on a calculator.

Angle 1 is arctan(horizontal/vertical) or, tan-1 (horizontal/vertical) on a calculator.

This assumes the third angle is 90 degrees.

[denrep]

Reduce the drawing to it's basics, a simple right triangle.

Of course, angle1 + angle2 (must) = 90 degrees

You can use a square and straight edge to find angles by reducing (or enlarging) your dimensions proportionately.

For a very simple example: Your dimensions are 12' x 12' at 90 degrees. Proportionately reduce your units of measurement to inches.
Lay a straight edge across a square from 12" to 12". A protractor or the "degrees obtained by using a square" chart, says this is 45deg. You know your total must be 90 deg, so subtract 45 from 90, your unknown angle is also 45 deg.

If you buy a high quality square, you can probably get a book with it. I have a book titled; Steel Square Pocket Book. It's crammed with unique (headache inducing!) measuring techniques and charts. This was common workaday stuff, for old time carpenters, and sailors!

Or determine as many knowns as you can, use an online plug in calculator!

Good Luck, let us know the angle of the dangle!

[tcbigair]

lay it out on paper and use a protactor

[Birdhunter1]

All,

The way I got it was " Some Crazy Thing - Oscar Has A Hairy Old A$$.

Sin = Opposite over Hypotenuse
Cos = Adjacent over Hypotenuse
Tan = Opposite over Adjacent



Oscar Had A Hold On Arthur is how I rembered it.

[yom_john]

mabe you could draw it out on paper scaled down and determine the angles
with a small protractor (dont no if that works)

[littlefuzz]

if you figure out one angle then the other would be 90 - whatever the other one was.

[William McCormick Jr]

http://www.Rockwelder.com/EastWood/nosine/nosine.html

This shows believe it or not once you are used to it, a rather easy formula to calculate one unknown angle of a right triangle. And of course then you have all three. But with no sin, cosine or tangent involved.

It takes a while to figure out what it is doing. But once you do you can use it from memory. No sine no scientific calculators. Just a regular old calculator with root function.

Sincerely,



William McCormick

[Tinbasher]

This seems like a very long way of doing things. You still use sine, cosine and tangent...you may not use the names, but in essence you find the ratios of the legs of a given triangle as the hypotenuse remains constant and describes the arc of a circle with a radius equal to its length . Sine ,cosine and tangent are just ratios and someone decided a long time ago that it would be easier to list them in a table as opposed to having to calculate them each time...which you have to do when using this method.

[denrep]

This seems like a very long way of doing things...

13162

[William McCormick Jr]

This seems like a very long way of doing things. You still use sine, cosine and tangent...you may not use the names, but in essence you find the ratios of the legs of a given triangle as the hypotenuse remains constant and describes the arc of a circle with a radius equal to its length . Sine ,cosine and tangent are just ratios and someone decided a long time ago that it would be easier to list them in a table as opposed to having to calculate them each time...which you have to do when using this method.

You are absolutely right about the ratios. That is all any formula dealing with a circle and angles is.

However, I meant that out in the field. When you do not have a calculator or computer handy. You can do it from scratch. I know one day I was out in the field and I wanted it bad.

I was looking at that formula, it could be stream lined a bit more. Made even more accurate. You could make it very easy to do. I just dreamed it up. But I checked it out pretty well and it is more then accurate enough for pipe bending and shop work.

I just put it up, so that others might just do that. Ha-ha.

I hate going to the book for sine values. Or using a calculator. And if I have my computer I use the Cadd program. Cadd is the fastest way. My Cadd program is like a giant calculator.

You seem to love math. I would love to run some of my teachings past you. Just to be able to bounce them off someone. Ha-ha.

I learned that Archimedes. Had built a giant stone, and polished it for years and years. Then rolled it on a flat polished stone. He according to my teachings thought pi was actually a little larger then 22/7. But in those days with no calculators, he thought that would suffice.

I actually machined a wheel out of 7075 T-8 aluminum tooling plate. I machined to a diameter just over 7 inches. The reason I stopped was that I had made a beautiful pass on the lathe. It looked like it was polished.

So I take it off put it on my polished 7075 T-8 aluminum table. I just gave the wheel a good wipe. In my mind if there were debris on it, it would roll a little farther, and then I would clean it and it would roll a lesser length in one revolution. I would measure it and that would be it.

I figured I would roll it and be able to show that pi was greater then 3.14159, so I roll it and it comes up about 3.14159 , so I get a little depressed. But I am a good sport. So I figured I would clean the wheel and roll it and report the roll with the best conditions. I clean it and it rolls further, it rolls to a ratio of 3.1430878987750706689998653923812

But in my mind when the wheel was dirty, I thought it was a little strange that it was rolling short and it was still dirty. I did not know that mathematically a dirty wheel does not go as far in one revolution.

If you ever have the time make a wheel and look how easy it is to measure a rolling object with a mark on it. Awesome experience. I use 3.14308 for my calculations nowadays. And it is an awesome number.

When you roll something around an object, you have to approximately use the center line of the object you are rolling. Meaning if it is 1/4 inch thick, you have to cut the material the length of a circle circumference for a circle diameter one quarter inch larger then the finished piece will be.

This is why often mathematicians get confused. Because they take a tape and roll it around an object. And use that as the measurement. But the tape shrinks on the inside and expands on the outside.



Sincerely,



William McCormick

[William McCormick Jr]

http://www.Rockwelder.com/WeldingWeb/tri/tri.html

This is a movie showing how we design and cut up big jobs.

This below is a movie about making pipe rails.

http://www.Rockwelder.com/EastWood/mathrail/mathrail.html

I can draw the simple ones up like that in about five minutes. The hard ones can take longer. When there are two views or multiple dimensions. Those are the exact shop drawings that create the rails.

http://www.Rockwelder.com/GeneralCadd/projects/PipeRails/Brownrails1.jpg
http://www.Rockwelder.com/GeneralCadd/projects/PipeRails/Brownrails2.jpg
http://www.Rockwelder.com/GeneralCadd/projects/PipeRails/Brownrails3.jpg
http://www.Rockwelder.com/GeneralCadd/projects/pr1.JPG
http://www.Rockwelder.com/GeneralCadd/projects/pr2.JPG
http://www.Rockwelder.com/GeneralCadd/projects/pr3.JPG
http://www.Rockwelder.com/GeneralCadd/projects/pr4.JPG



Sincerely,



William McCormick

[Tinbasher]

Thanks for sharing. I've always loved math...it doesn't lie. Regardless of which of the thousand methods you use to solve a problem...you always arrive at the same answer.

Mean diameter is a concept that I use on a daily basis. I use decimal equivalents for different gages and calculate ID's, OD's and wrapper lengths...they are always deadly accurate.

The formula that we were discussing earlier that relates angle, radius and chord length is also very handy. I like to use it for radial layout of cones. I have a formula that lets me determine the angle of a given pattern, and then from that angle I calculate the chord length and I can make the pattern without bending my 4 foot circumference rule into an arc and trying to measure the arc length.

Here's a neat one for wrapper length...

Wrapper Length= sin 1 degree x radius x angle.

All this being said, a good mechanic has to know when to put his calculator away and work practically. I would never calculate a bend allowance for braking. It is much easier to put a piece of scrap in there and stomp on the pedal. Know what you started with and then by measuring what you finish with...you can quickly determine exactly what happened....the same for bending tubing I'll assume.( something that I have never done )

[William McCormick Jr]

Thanks for sharing. I've always loved math...it doesn't lie. Regardless of which of the thousand methods you use to solve a problem...you always arrive at the same answer.

Mean diameter is a concept that I use on a daily basis. I use decimal equivalents for different gages and calculate ID's, OD's and wrapper lengths...they are always deadly accurate.

The formula that we were discussing earlier that relates angle, radius and chord length is also very handy. I like to use it for radial layout of cones. I have a formula that lets me determine the angle of a given pattern, and then from that angle I calculate the chord length and I can make the pattern without bending my 4 foot circumference rule into an arc and trying to measure the arc length.

Here's a neat one for wrapper length...

Wrapper Length= sin 1 degree x radius x angle.

All this being said, a good mechanic has to know when to put his calculator away and work practically. I would never calculate a bend allowance for braking. It is much easier to put a piece of scrap in there and stomp on the pedal. Know what you started with and then by measuring what you finish with...you can quickly determine exactly what happened....the same for bending tubing I'll assume.( something that I have never done )


Is there a friendly diameter? I have enough mean stuff already. Ha-ha. By mean you mean average diameter?

Wow, I will go check out that wrapper length formula.

For break bending you can actually figure it out ahead of time. Most sheet metal stuff is done with a sharp die, no radius.

So all you have to do is create the inside measurements of the duct, and mark them on the sheet and know that the outside will increase by the thickness of the metal all the way around.

You are probably all to aware of this, I just know that when I was a kid I ruined my last piece of metal not knowing this. And I had no way to cut it after it was bent. I am sure there is some kid around here that does not need to do that.

With thicker material the same is true. If you take a 4"x4" piece of 1/4" metal and bend it in the center to ninety degrees, with a sharp knife. The inside will measure 2"x2" and the outside will measure 2 1/4" x 2 1/4". Closest thing to free metal. Ha-ha.

http://www.Rockwelder.com/Flash/breakwave/breakwave.html

My father and guys from Grumman could only imagine the world I was going out into. So they really went out of their way to make sure I heard about every factory accident. From welding machines rocking back and forth, with thunderous crashes. To guys getting killed by pencils and high voltage.

This link above, my father warned me about the principle, when I was a kid. He said that metal can form a wave. And this wave has ability to deliver energy not expected.

He told me never put my hands under a sheet of metal, when you engage the shear. Or break. The reason is that a tool not seen under the metal could cause the metal to create a wave as the hold downs or blade come down to cut. If your hand is between the metal the side stop, you could loose a finger or two, with a heavy sheet of metal.

The union sheet metal guys know not to carry sheet metal over their heads for this reason. And will yell at you if you do it. You can break a neck.

For bending pipe you do need a couple test pieces. Once you have them, you just alter the macro in the Cadd program. What I do is I substitute a different formula for Pi. If I want really accurate bends. I use 2.93 I believe.

I usually find that my pipes come out a quarter inch longer if two nineties are created in one length, using standard pi. So I just reduce pi to create a slightly shorter output.

Sincerely,



William McCormick

[William McCormick Jr]

I also use the center line radius of the pipe. To come up with the bend marks. And cut marks. But alter that just a bit with the slightly smaller pi formula.


Sincerely,



William McCormick

[William McCormick Jr]

Thanks for sharing. I've always loved math...it doesn't lie. Regardless of which of the thousand methods you use to solve a problem...you always arrive at the same answer.


See I started talking to you and already I am getting great ideas.

You can just use half the difference, in the length of the hypotenuse, and the long leg of a right triangle to create the radius of a circle.

To come up with the circumference of that circle halfway between the two circles. Add the difference between the hypotenuse and the long leg, to the circumference.

Divide that by the short side of the triangle, and divide that number into 360 for degrees to get the angle of the triangle.

Still not amazingly short, but possible in the field. And you don't need a calculator for that.

What I really want and almost lost my mind trying to find years ago. Was an easy way to calculate the Root square of a number. Without a calculator.

Once I was close, but I could not get the ratio right. I believe that by squaring the number you want the root square of. You can create a ratio back to the square root of the original number. But until I do, I need a calculator. Even in programming on IBM computers they loop the square root formula until its right.

Sincerely,



William McCormick

[William McCormick Jr]

13162


How do you use that thing?

I used to use pitch to cut the first rafter. A friend of mine and sometimes partner in building actually showed me how to do it. They make little brass guides you can lock onto the square, so you can just move it up the rafter.

But I don't get how you are going to get degrees from that. I would love to see how they do it.


Sincerely,



William McCormick

[denrep]

How do you use that thing? ...

"That thing" probably holds the secret to the universe!

The degrees are obtained, by referencing measurements to the "degrees obtained by using the square" chart.

Steel Square Pocket Book by Dwight Stoddard, covers in depth, uses of the square.

Besides the usual builder's uses, Dwight Stoddard wrote in detail how to use the square to lay out angles, circles, ellipses, hexagons, and seven other "agons". Also covered are things like using the square for a calculator, long distance sight measuring tool, drawing resizing tool, etc. He writes that the size of the book, limits it to mentioning just some of the square's uses! I guess he skipped paint mixer and can opener!

Of course, everyone remembers these: 13196

[William McCormick Jr]

Here's a neat one for wrapper length...

Wrapper Length= sin 1 degree x radius x angle.


Tried that out very smooth with a scientific calculator.



Sincerely,



William McCormick

[denrep]

Okay mathemagicians, Back to post one. Forget the angles. Is there a simple way to come up with two lengths for the brace? One for the inside one for the outside, with the inside length being known.

Besides measuring it! :realmad:

[Weldordie]

Okay mathemagicians, Back to post one. Forget the angles. Is there a simple way to come up with two lengths for the brace? One for the inside one for the outside, with the inside length being known.

Besides measuring it! :realmad:

Are we building a watch or welding iron?

[denrep]

Are we building a watch or welding iron?

A welding iron, but we have to deliver it to Mars, ready to install!

[Weldordie]

A welding iron, but we have to deliver it to Mars, ready to install!

Okay. Now I get it.:rolleyes:

[Brad Blazer]

The outside edge length(OL) of the brace will be the inside length(IL) plus thickness(T) times the sum of the tangent(tan) and cotangent(cot) of either angle. :sleeping:

OL = IL+ T(tan+1/tan)

(cot = 1/tan)

A spreadsheet is the best calculator on the computer even if it's MSWorks.:jester:

[Weldordie]

The outside edge length(OL) of the brace will be the inside length(IL) plus thickness(T) times the sum of the tangent(tan) and cotangent(cot) of either angle. :sleeping:

OL = IL+ T(tan+1/tan)

(cot = 1/tan)

A spreadsheet is the best calculator on the computer even if it's MSWorks.:jester:

Those Martians will be soooo happy!:blob4:

[denrep]

The outside edge length(OL) of the brace will be the inside length(IL) plus thickness(T) times the sum of the tangent(tan) and cotangent(cot) of either angle. :sleeping:

OL = IL+ T(tan+1/tan)

(cot = 1/tan)

A spreadsheet is the best calculator on the computer even if it's MSWorks.:jester:

Thank you, Has anyone run it yet?

Now, if that works, one more number and we're done.

We'll presume that "T" (brace Thickness) is 10.

13305

How do we align Outside Line Center, with ILC to make sure that IL and OL end up correctly oriented to each other.

This is not a trick question. I don't know the answer, but keep thinking that there must be a simple mathematical way to figure the brace.

[Brad Blazer]

The angles are 53.13 and 36.87 deg.
If we pick 53.13, Tan = 4/3, cot = 3/4 OL = 50+ 10(4/3+3/4) = 70.83
The upper ? = 10 x 5/4 = 12.5
The lower ? = 10x5/3 = 16.67
The triangle to the outside of the brace has sides 42.5, 56.67, 70.83 which follow the same 3-4-5 ratio.
A line from the 90 will pass thru ILC and OLC. OLC is offset from ILC by 10(4/3 - 3/4) or by 5.83.

If I was cutting the brace I would miter 36.87, measure 50 along the inside edge, then miter 53.13.

[aDreamSoReal]

the easy way....

http://www.boschtools.com/tools/tools-detail.htm?H=193849&G=80379&I=55003

[William McCormick Jr]

A welding iron, but we have to deliver it to Mars, ready to install!

If I had to cut up 1"x1" box tubing. I would create a giant square with two pieces of the box tubing, 1 1/4" off the floor, using blocks here and there.

Using a 3,4,5 triangle or 6,8,10 or what ever it took. And then just move the angular piece to where it was supposed to be, mark the 1"x1" tubing with some soap stone, and cut it.

What are you going to cut that acute angle with anyway. A hand held hacksaw? Or a hand held circular saw? Neither of which will benefit from a need of more accuracy then just laying it out.

Also if you are welding, the inside of those angles will pull in and warp all the 1"x1" tubing anyway. The inside weld is much stronger.

Sincerely,



William McCormick