American Welding Society Forum

Looking for some help figuring out how to calculate distance equivalent to degrees.

e.g.: If I have a 2 degree tolerance for perpendicularity at 20" from the joint how much is 2 degrees in inches?

how would I figure out the given tolerance in inches over a given distance.

ex: 2 degrees at 20", 4 degrees at 20 inches etc.

Thank you everyone for your help

e.g.: If I have a 2 degree tolerance for perpendicularity at 20" from the joint how much is 2 degrees in inches?

how would I figure out the given tolerance in inches over a given distance.

ex: 2 degrees at 20", 4 degrees at 20 inches etc.

Thank you everyone for your help

The easiest way to explain, would be to take the radius (20"), multiply by two, to get your diameter. Multiply that by pi, divide by 360, and multiply that by your (2) degree tolerance.

edit: I still stand by my answer if your on a shop floor and don't have a scientific calculator, and when you need an answer quick.

you will be rounding a couple thousandths anyway.

edit: I still stand by my answer if your on a shop floor and don't have a scientific calculator, and when you need an answer quick.

you will be rounding a couple thousandths anyway.

Drew,

If I understand your question you are trying to determine how far "out of plumb" or out of perpendicular a line would be if it was 2 degrees over 20" and/or 4 degrees over 20". If this is the solution you are looking for it would be solved with right triangle trigonometry. You need to know either 2 legs of a right triangle or 1 leg and one angle [excluding the 90 degree angle] of the right triangle to determine the other angles and leg lengths. While a fairly easy solution to work through it is beyond an easy explanation here and I don't think you would be best served for someone to just give you the answer. The question you ask [if I am understanding it correctly] is a calculation you will need to be able to solve for 10,000 times in the course of your trade work so it behooves you to understand how to do that. There are calculators online that will just give you the answer but again the answer is not what you need it is the method to find the answer. It's called Right Angle Trigonometry.

The very best text book I have found to explain how this is done is Johnny E. Hamilton's: Pipe Fitters Math Guide. This book starts you at the very beginning steps of the necessary arithmetic reviews and examples in short clearly explained chapters each with its own quiz and the answers for review in the back so you can test yourself as you learn. It progresses step by step to build on the foundation to teach you how to use right angle trigonometry to solve the problem you've posed. I have recommend this book to many men I've worked with over the years as an absolute fundamental and necessary bank of knowledge they need to have in their understanding. Those who ignored that recommendation never really got very good at building things. Those who did had doors of understanding opened for them that helped to build their lives for the better. Knowledge is Power. You need the knowledge that is in this book. The book only costs 25 dollars. The knowledge you'll gain will be impossible to put a price on. The effort to acquire that knowledge will be an investment in yourself that will accrue interest for the rest of your life.

Best of luck.

If I understand your question you are trying to determine how far "out of plumb" or out of perpendicular a line would be if it was 2 degrees over 20" and/or 4 degrees over 20". If this is the solution you are looking for it would be solved with right triangle trigonometry. You need to know either 2 legs of a right triangle or 1 leg and one angle [excluding the 90 degree angle] of the right triangle to determine the other angles and leg lengths. While a fairly easy solution to work through it is beyond an easy explanation here and I don't think you would be best served for someone to just give you the answer. The question you ask [if I am understanding it correctly] is a calculation you will need to be able to solve for 10,000 times in the course of your trade work so it behooves you to understand how to do that. There are calculators online that will just give you the answer but again the answer is not what you need it is the method to find the answer. It's called Right Angle Trigonometry.

The very best text book I have found to explain how this is done is Johnny E. Hamilton's: Pipe Fitters Math Guide. This book starts you at the very beginning steps of the necessary arithmetic reviews and examples in short clearly explained chapters each with its own quiz and the answers for review in the back so you can test yourself as you learn. It progresses step by step to build on the foundation to teach you how to use right angle trigonometry to solve the problem you've posed. I have recommend this book to many men I've worked with over the years as an absolute fundamental and necessary bank of knowledge they need to have in their understanding. Those who ignored that recommendation never really got very good at building things. Those who did had doors of understanding opened for them that helped to build their lives for the better. Knowledge is Power. You need the knowledge that is in this book. The book only costs 25 dollars. The knowledge you'll gain will be impossible to put a price on. The effort to acquire that knowledge will be an investment in yourself that will accrue interest for the rest of your life.

Best of luck.

Hello Drew, I might as well throw something in here to "confuse" you further, no, just kidding. I'll see about attaching a simplified trig. chart that I use in a fabrication class that I teach.

"Simplified" when the hypotenuse is known and one angle of the right triangle besides the 90 degree corner is also known. Then you can use that information to figure out the lengths of the other unknown sides with the use of pythagorean theorem.

The same can be said when you know one of the other sides(side adjacent, side opposite) of the right triangle and one angle, then you can use tangent and pythagorean theorem once again to figure out the lengths of the unknown sides of a right triangle.

One other handy rule to keep in mind is that any triangle contains 180 degrees and a right triangle always contains one 90 degree corner and if you know the angle of one of the other "corners" then you can easily solve for the unknown one.

"Simplified" when the hypotenuse is known and one angle of the right triangle besides the 90 degree corner is also known. Then you can use that information to figure out the lengths of the other unknown sides with the use of pythagorean theorem.

The same can be said when you know one of the other sides(side adjacent, side opposite) of the right triangle and one angle, then you can use tangent and pythagorean theorem once again to figure out the lengths of the unknown sides of a right triangle.

One other handy rule to keep in mind is that any triangle contains 180 degrees and a right triangle always contains one 90 degree corner and if you know the angle of one of the other "corners" then you can easily solve for the unknown one.

Attachment: Trigtangentandsinefunctions.docx (36k)

miracle point (angle-pitch) 2 degrees =.035 per inch .035x20= .7inch, 4 degrees =.07 per inch .07x20=1.4 inch

1 degree angle = .0175 pitch per inch or .210 inch per foot

1 degree angle = .0175 pitch per inch or .210 inch per foot

Dick,

*WELCOME TO THE AWS WELDING FORUM!!*

Now that's something I can work with, keep it simple. Nice first post.

He Is In Control, Have a Great Day, Brent

Now that's something I can work with, keep it simple. Nice first post.

He Is In Control, Have a Great Day, Brent

Thanks Brent, this was in responce to Drew's post. My post didn't go where I intended.

Dick

Dick

I knew that and that is what I was referring to, simple methods of finding how far off of plumb. Like using 3/4/5 instead of square roots and all the rest of it for finding if an object is square.

Brent

Brent

b=(20 x sin B)/ sin 90 where B = 2 degrees (or the appropriate tolerance) = 0.697989934 inch

b = (20 x 0.034899497) / 1 = 0.697989934

convert to 16ths = 11/16 inch. So, the 2 degree tolerance for the 20 inch length is 11/16 inch to the left or right of plumb.

The law of sins was the rule used to solve the problem

a/sin A = b/Sin B = c/Sin C

where the sin of 90 degrees is equal to 1 and the values of a, b, and c are the dimensions of the right triangle where one side (a) is the radius of the circle (the longest side)

I use a TI30 to solve the equations.

Best regards - Al

b = (20 x 0.034899497) / 1 = 0.697989934

convert to 16ths = 11/16 inch. So, the 2 degree tolerance for the 20 inch length is 11/16 inch to the left or right of plumb.

The law of sins was the rule used to solve the problem

a/sin A = b/Sin B = c/Sin C

where the sin of 90 degrees is equal to 1 and the values of a, b, and c are the dimensions of the right triangle where one side (a) is the radius of the circle (the longest side)

I use a TI30 to solve the equations.

Best regards - Al

Al is exactly correct.

A simpler way ( for me ) is to use the Trig formula Hypotenuse = Opposite divided by the Sin of the angle.

Using Al's drawing the longer red line is the hypotenuse, 20 inches, and the short red line is the opposite, what we want to know. I believe Al just used 25 on his drawing as an example. Your case you need to know what 2 degrees on 20 is.

Using algebra, we can change the formula to Opp= Hyp times Sin of 2 degrees.

Use a scientific calculator or look up on a trig chart to find the Sin of 2 degree or 4 degrees or what ever.

So 20 inches times 0.0349 (sin of 2 degrees) = 0.698 inches about 11/16

20 times 0.698 (sin of 4 degrees) = 1.396 inches, a little over 1 3/8.

Written out it seems much more confusing than it really is. Learning and understanding basic Trig has been invaluable in my fabrication career.

2 degrees seems like a very generous tolerance on plumb.

Keep the bubble between the lines and you will be golden.

Floyd

A simpler way ( for me ) is to use the Trig formula Hypotenuse = Opposite divided by the Sin of the angle.

Using Al's drawing the longer red line is the hypotenuse, 20 inches, and the short red line is the opposite, what we want to know. I believe Al just used 25 on his drawing as an example. Your case you need to know what 2 degrees on 20 is.

Using algebra, we can change the formula to Opp= Hyp times Sin of 2 degrees.

Use a scientific calculator or look up on a trig chart to find the Sin of 2 degree or 4 degrees or what ever.

So 20 inches times 0.0349 (sin of 2 degrees) = 0.698 inches about 11/16

20 times 0.698 (sin of 4 degrees) = 1.396 inches, a little over 1 3/8.

Written out it seems much more confusing than it really is. Learning and understanding basic Trig has been invaluable in my fabrication career.

2 degrees seems like a very generous tolerance on plumb.

Keep the bubble between the lines and you will be golden.

Floyd

You are right Floyd, the 25 inches was used as an example. I used 45 degrees because one can see the relationship between the 25 inch radius and the 17+ dimension.

Then I used the same equations to solve for 20 inches. The radius of the circle is always the known value in my examples. The tolerance is the vertical dimension. In this case it is about 11/16 inch.

The relationship is called the Sin Law or Law of the Sins. It is pretty easy to remember. But most of us need a calculator to do the calculations.

I tried to use Excel, but that only produces answers in terms of radians.

Al

Then I used the same equations to solve for 20 inches. The radius of the circle is always the known value in my examples. The tolerance is the vertical dimension. In this case it is about 11/16 inch.

The relationship is called the Sin Law or Law of the Sins. It is pretty easy to remember. But most of us need a calculator to do the calculations.

I tried to use Excel, but that only produces answers in terms of radians.

Al

It's short for: "

Some more:

And:

There are probably more representations but, I'm going to stop here...

No seriously folks, this acronym stands for:

I'm not going to explain anymore than what has already been shown to you and there's no need to be redundant here... The only advice I have to give is if you're still not confident enough to use what has been described to you, then here's a link which in itself is easy enough to understand that will help you understand better how to solve triangle problems. I would also second Floyd's recommendation on buying the book he endorsed because it is a well written one for a student to become more comfortable and confident in using trigonometry to find missing dimensions whether they be linear or angular...

I would also recommend this link to use for learning how to become more familiar with Trigonometry and for practice solving triangles as well as learn more about the 3 main functions and the secondary functions even if they're not used as often...

This link offers tips for example on how to "Solve" triangles... "Solving" means finding missing sides and angles... When we know any 3 of the sides or angles... We can find the other 3*... *( Except for 3 angles, because we need at least one side to find how big the triangle is.) As well as other tips to help you become more comfortable with Trigonometry:

http://www.mathsisfun.com/algebra/trig-solving-triangles.html

This link will also offer Triangle Solving Practice in another link which can be found in the link I just posted... The bottom line is this, it's one thing to solve the problems by pushing buttons on a calculator and another to understand what, why and how to solve problems where trigonometry is concerned like Floyd also mentioned... Finally, Al's post is very descriptive and very useful for you also to familiarize yourself further with the methods used to solve a variety of triangular problems... Once you get used to using the various functions and applying the tips given, you should not have any problems finding missing angles and sides. Enjoy!

Respectfully,

Henry

Funny you should mention trippin on acid. I watched a great documentary on acid the other night. I’ve found that it's a pretty cool way to watch great documentaries. They said that the hallucinogenic effects of acid were discovered in 1943. At least that’s what the dragon with a megaphone told me. It was also mentioned in the documentary that acid helps alcoholics give up drinking. My analogy of that statement is that yeah, it’s in the same way that a bullet helps you give up oxygen. The fact is that you can stop doing drugs if you put your mind to it. I was able to stay away from cocaine after I bought a twelve foot straw. Staying away from acid has been a little different. My doctor said that over the years, acid has had a serious effect on my mental health and has altered my state of mind to the point where I'm walking around in a constant dream world. I didn’t like his evaluation so I decided to get a second opinion. I found out that he was overreacting because the Lizard King from the 5th dimension of Nazkabar said that I was just suffering from stress. My best friend actually supported what my doctor had told me. He said that I’d been taking too much acid, but c’mon, what does a dog know about taking drugs? We all know that if everybody in sports did drugs, it would be a level playing field, unless the drug was acid. Then it would change from a level playing field to a castle on a hill made of lava. That thought came to me one night after I hallucinated that I was a very well educated giant. And they say taking drugs don't make you big or clever. The only time LSD hasn’t worked for me is when I’ve tried to use it as a word in a game of strip Scrabble. And those of you who think LSD is dangerous have obviously never made sweet love to a unicorn. It was just a one night stand. I was at a club one night and I needed a ride home. The unicorn evidently had other plans and he wanted to see me again. They’re very strange creatures and they’re extremely clingy. It’s really pretty simple logic. A dog says “woof woof”. A cat says “meow”. A cow says “mooooo”. But a unicorn says “Where have you been?! I've been looking everywhere for you. You were supposed to be here at 7 o’clock to meet my parents. You've ruined everything. What kind of sick game are you playing!?! Anyway, a word of advice about tripping. Even though they help people plan and book the perfect trip, I don’t recommend going to the Trip Advisor website looking for their recommendations for some good acid or magic mushrooms, and then complaining to management when you don’t get it. The only thing I get, is, when life hands you lemons, stop taking LSD.

Dear All,

Greetings!

I hope this mail finds you well and good.

I would need a big favor from you. I have come across with a new work in the job of rail welding. I am not very much skilled in this procedure. Could I please ask you to share the procedure if you have one with you? If not, please tell me what code or references clause I should follow. And also if I have completed work with the procedure and a repair is found please refer me the references that tell me the Repair procedures. Thank you, my dear friend, for your valuable help in this regard.

Note : If any one Please send me the Rail welding procedure (PQR + WPS) and repair welding procedure if you have. I need that to refer some requirements.

Welding procedure: Thermit welding

Raja Magesh

Greetings!

I hope this mail finds you well and good.

I would need a big favor from you. I have come across with a new work in the job of rail welding. I am not very much skilled in this procedure. Could I please ask you to share the procedure if you have one with you? If not, please tell me what code or references clause I should follow. And also if I have completed work with the procedure and a repair is found please refer me the references that tell me the Repair procedures. Thank you, my dear friend, for your valuable help in this regard.

Note : If any one Please send me the Rail welding procedure (PQR + WPS) and repair welding procedure if you have. I need that to refer some requirements.

Welding procedure: Thermit welding

Raja Magesh

There was some great math on this thread, and of course Scottn's usual insight, but I have to mention that there are now cheap digital levels that will, at the push of a button, make eighth grade unnecessary.

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