# Trig Question



## jocat54 (Jan 29, 2017)

I have a question for those of you who know trig.
When I use an online calculator for sine bar height for an angle of 20 degrees and a sine bar length of 2.5" I get an answer of 0.8551 inches---but when using windows scientific calaculator and using the formula, Tangent A x b, I get .9099255xxxxxxx. See highlighted below. What am I doing wrong?


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## Bill Gruby (Jan 29, 2017)

If Angle A is 20*, its Tangent is .363970. That times "b" (2.5) is .909925.

"b" (2.5) / cotangent A (2.747477)  is 909925

 Your answer is .909925

 I did it both by the book and tables and long hand.

 "Billy G"


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## Tony Wells (Jan 29, 2017)

Unless I am misunderstanding something about what you are asking, "c", the hypotenuse, is the 2.5000" constant for your sine bar, not the base of the triangle "b". As you elevate the sine bar, "c" stays the same and is used for calculations, whereas "b" will get shorter.  So your gage block stack (typically used) would be sin20°*2.500

But it's been a long day and maybe I am misunderstanding something.


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## Bill Gruby (Jan 29, 2017)

I gave him the right one of both his answers.
I agree the side "c" is the 2.5 side.
.342020 X 2.5 =  85505


 "Billy G"


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## Tony Wells (Jan 29, 2017)

Umm, but the gage block stack for a 2.500" sine bar at 20° would be 0.8551.

Side "b" is unknown at this point. The only constants are side "c" (2.5000) and the desired angle of 20°. Therefore you cannot use side "b" for your stack calculations.


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## Bill Gruby (Jan 29, 2017)

look again LOL I was editing while you answered. You actually have 2 angles not one. The 90*. So you actually have all three angles. 90 + 20 + 70 = 180  LOL You are correct though, side "b" is of no concern here.

 "Billy G"


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## Tony Wells (Jan 29, 2017)

Correct, that 90° is assumed since we are dealing only with a right triangle. That's what makes everything else work.


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## Bill Gruby (Jan 29, 2017)

I used to be much better at the math game but age has taken its toll on the gray matter.

 "Billy G"


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## Tony Wells (Jan 29, 2017)

That it does Bill.....that it does.


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## rdean (Jan 29, 2017)

I never got very far doing trig so for me the easiest way to find the stack is use the Machinery Hand book.
Look up a 20 degree angle in the table and divide by 2 since the sign bar John is using is half as long as in the table.
1.71010 divided by 2 = 0.85505 rounded up to 0.8551.

I don't have to try and remember the formulas. 

Now if I could just find my glasses.     
Ray


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## Bill Gruby (Jan 29, 2017)

As long as you get the correct answer you did it right.  You got it right.  

 "Billy G"


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## Wreck™Wreck (Jan 29, 2017)

Edited to not give incorrect data.


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## Tony Wells (Jan 29, 2017)

Wrong answer, Wreck. Use aligned dimension. The hypotenuse is 2.5oo, not the base.

It's simple. SinA*c-c of rolls give you stack height.


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## jocat54 (Jan 29, 2017)

Thanks for all the replies.
The base (side b) is 2.5" and the angle is 20 degrees.
I think the right answer is .909925 by the formula--I guess I am just missing on how to use the online calculators (putting in the wrong numbers?)

edit: I think it just clicked for me after reading Tony's 1st post again.


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## T Bredehoft (Jan 29, 2017)

Hmmm.  My 20 year old TI36X calculator returns for 2.5 x sin 20 = 0.855050358. That's the stack of Jo blocks. 2.5 has gotta be the hypotenuse.


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## Wreck™Wreck (Jan 29, 2017)

Gunrunner you need to order a set of these for the Z axis hand crank. Bill did you just spit coffee on your monitor? :rofl:

View attachment 253356


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## Tony Wells (Jan 29, 2017)

Right you are Tom.


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## Tony Wells (Jan 29, 2017)

John, side "b" is totally immaterial. When you sit the sine bar on a surface plate, for example, it may as well be infinitely long. What changes, as Wreck illustrated, is the distance from the bottom of the stationary roll to the roll that is elevated. In the plane represented by the surface plate in my illustration, the distance between the rolls "appears" to get shorter, the only option it has when you rotate it out of a plane parallel with the reference (surface plate or vise, or machine table, etc)BUT, the roll c-c stays the same, no matter what. That's the value you have to know to calculate the stack, which of course varies according to the angle. You can safely ignore the length of side "b".


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## jocat54 (Jan 29, 2017)

Thanks all--I THINK I finally understand.


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## Tony Wells (Jan 29, 2017)

Do the calculations for a 45° angle, and set your protractor up against it, or a 45° angle block. At that severe an angle, the error will be obvious if you use 2.500 for side "b" in the calculations. Then set it using 2.500 as side "c" (hyp) for sin multiplier. It will then match your protractor.

Use the formulas on the second line in your chart. Angle A and side "c" are your knowns.


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## Reeltor (Jan 29, 2017)

The programing in the calculator may be wrong.  I think it was This Old Tony who recently posted on a YouTube video that he has had calculators and on-line computer programs give incorrect answers.  Tony or whoever it was, said it was rounding errors and suggested to use tried and true tables.


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## jocat54 (Jan 29, 2017)

If I am understanding right I could use the chart calculations if I wanted to drill 2 holes in a plate that were at a 20*---but using a sine bar the same formula would be wrong because of the arc of the sine bar centers. Don't know if I explained that very well but I do understand now. Thanks to all


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## Tony Wells (Jan 30, 2017)

Well, not exactly.....

In the case of drilling 2 holes, let's say that they line described by the center of each holes lies at 20° from the hard jaw of your vise, for example......

If you are using Cartesian coordinates, the first hole wold be considered the origin, in all likelihood, even if it were an inch off the hard jaw and half an inch from the edge of the material. It's simpler to explain if we set our X and Y zeroes at the first hole. Let's say that your hard jaw represents the base of that triangle, end hence the "c" figure. We've determined that the length of side "c" is immaterial. It's just a reference line (or plane if you want to think about all this in 3D, and it gets a bit hairy then). So you can move along the X axis any desired distance. But as of yet you don't know that distance. But you know the desired angle to be 20°. IF you wanted the holes spaced 2.500" apart c-c, then we can see how this works like the sine bar. Your DRO won't tell you the length of an angular move, directly. But we know it will be less than 2.500" because of the arc swept by what now pretty much amounts to a piece of a bolt circle. We know the -Y- move, however, to be 0.8551. So now we have information to use another section of your chart. We know side "a"(0.8551), we know angle "A" (20°), and we know side "c", (2.500)

So our second hole will be at -Y-0.8551 x -X-2.3492

cosine A * c = side B, which is the -X- axis move.


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## CluelessNewB (Jan 30, 2017)

Remember SOHCAHTOA  

Sin = Opposite/Hypotenuse 
Cos = Adjacent/Hypotenuse
Tan = Opposite/Adjacent

We know the Hypotenuse (2.5") and we know the angle (20 degrees), we are looking for the opposite side. 

The equation we want is the one that has the 3 things we either know or are interested in knowing and that would 
be: Sin = Opposite/Hypotenuse 

Rearrange the equation so that we get the unknown (Opposite) by itself:
To do that we multiply each side of the equation by the Hypotenuse:

Hypotenuse x Sin = Hypotenuse x Opposite / Hypotenuse

The two "Hypotenuse" terms on the right side cancel out so this reduces to:

Hypotenuse x Sin = Opposite

swapping sides to make it look nicer:

Opposite = Hypotenuse x Sin

Plug in the stuff we know:

Opposite = 2.5 x Sin(20)

Opposite = 2.5 x 0.34202

Opposite = 0.85505


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## willthedancer (Jan 30, 2017)

I posted about this in 'Bad Math' elsewhere. Use the table rather than the calculator.

Any angle on the sine bar will work out because the rolls are the same diameter, so the tangent points will always be the length of the sine bar(hypotenuse). Just look up the sine of the angle you want, and multiply it by the length of the sine bar, and you have it. Adjustable parallels are handy here too, and excuse you from having to buy expensive Jo Blocks

Rich's calculations above are right on!


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## davidpbest (Feb 4, 2017)




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## jamby (Feb 4, 2017)

That's why they call it a "sine" bar, or sine plate.  Use the sine of the desired angle  (20 deg) times the length of the bar.    .342 x 2.5 = .855


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## easymike29 (Feb 4, 2017)

This might make it a little clearer.


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## jamby (Feb 5, 2017)

Well I'll try to give an answer to the original question with it a bit more clarity.  The formula tan A x b is a trig function to derive the length of side a.  Which is a fine bit of math but useless for what you want with a sine bar.   Sine bars and sine plates are built to take advantage of the rule that the side opposite in a right triangle is the sine of the angle.  No trig function is required to determine what size block to use, only the know angles sine x the length of the bar 
Easymike nice drawing.  Looks much like the sine bar I made for myself 50 years ago.


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## CluelessNewB (Feb 5, 2017)

jamby said:


> No trig function is required to determine what size block to use, only the know angles sine x the length of the bar



You are using a trig function, that is what "sine" is!   You don't need to derive the equation each time, I was just trying to show above where it came from.


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