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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Trig Question
- Thread starter jocat54
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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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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.
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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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
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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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!
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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This might make it a little clearer.
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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.
Easymike nice drawing. Looks much like the sine bar I made for myself 50 years ago.
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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.