How do you calculate this?

MazatrolMatrix · Oct 15, 2021

So I want to know how to calculate the distance from the center perpendicular to the plane that is 40 degrees from centerline. The computer tells me it's approx 61.799 mm, but I'd really like to know how to it by hand. Hopefully someone can help

Edit: oops sorry about the sketch there

View attachment 332213 View attachment 332214

mhajicek · Oct 15, 2021

You could trig it out, but it would be a pain and prone to error. Snap a dimension on the sketch on the computer and read the result.

mhajicek · Oct 15, 2021

Just sketched it in Mastercam, dimension says "61.800".

MazatrolMatrix · Oct 15, 2021

Yeah I did that, got the same result as you.

I tried to trig it out myself but I couldn't do it, which I found frustrating. Or frankly, I managed to calculate it but I didn't understand how I did it, the logic behind it, just arrived at the right conclusion somehow since I already knew the answer. Would really like to learn how to solve it by hand.

wheelieking71 · Oct 15, 2021

MazatrolMatrix said:
Yeah I did that, got the same result as you.

I tried to trig it out myself but I couldn't do it, which I found frustrating. Or frankly, I managed to calculate it but I didn't understand how I did it, the logic behind it, just arrived at the right conclusion somehow since I already knew the answer. Would really like to learn how to solve it by hand.

Your hankerin for smarts is admirable! But don't feel bad, lots of us rely on the magic pixie thinking boxes for this stuff. I know I do!
I was a straight A student in math and geo/trig in high-school (34 years ago). I could have figured that out lickity=split back then.
Now? Nope! I haven't used those cerebral compartments in so long, the lights wont turn on any more.

I work in imperiel, so I converted your numbers to the 5th decimal, and came up with 61.79832

gregormarwick · Oct 15, 2021

extend the 40° line to where it intersects with the centre line by constructing a triangle from the 38mm flat.

You then are left with a single triangle that you can calculate the length you require.

MazatrolMatrix · Oct 15, 2021

wheelieking71 said:
Your hankerin for smarts is admirable! But don't feel bad, lots of us rely on the magic pixie thinking boxes for this stuff. I know I do!
I was a straight A student in math and geo/trig in high-school (34 years ago). I could have figured that out lickity=split back then.
Now? Nope! I haven't used those cerebral compartments in so long, the lights wont turn on any more.

I work in imperiel, so I converted your numbers to the 5th decimal, and came up with 61.79832

Haha, thanks. Well, I really think there's value in trying to calculate things yourself instead of just relying on software all the time. Sort of like studying for a test and memorize the questions, maybe you pass but did you really understand it, did you really learn anything. If anything it feels good to know how things are done!

MazatrolMatrix · Oct 15, 2021

gregormarwick said:
extend the 40° line to where it intersects with the centre line by constructing a triangle from the 38mm flat.

You then are left with a single triangle that you can calculate the length you require.

Thanks alot Gregor!

gregormarwick · Oct 15, 2021

MazatrolMatrix said:
Thanks alot Gregor!

You're welcome.

Note that is only one way. You could just as easily split the four sided area up into two triangles with a line from the centre to the corner of the 38mm flat and do it that way. Basically any way that you can find to split it up with two right angle triangles and you're basically done.

Bobw · Oct 15, 2021

gregormarwick said:
You're welcome.

Note that is only one way. You could just as easily split the four sided area up into two triangles with a line from the centre to the corner of the 38mm flat and do it that way. Basically any way that you can find to split it up with two right angle triangles and you're basically done.

That's how I was going to do it at first, but when I started my little sketch, I changed my mind..
Just because.. Not because one way is better than the other.

And yeah.. Solve one triangle (that you have to make up).. some addition, and that gives you the hypotenuse
on your second triangle, and that's your answer.. I got 61.79973373.

I enjoy these threads.. Every once in a while its nice to find out I still remember how to do the math.

MazatrolMatrix · Oct 15, 2021

gregormarwick said:
You're welcome.

Note that is only one way. You could just as easily split the four sided area up into two triangles with a line from the centre to the corner of the 38mm flat and do it that way. Basically any way that you can find to split it up with two right angle triangles and you're basically done.

I tried to calculate it but I ended up with 68,82. The correct answer is 61,799. Must still be something missing!

Oh, never mind! :scratchchin:

Orange Vise · Oct 15, 2021

Look at the skinny triangle first. 19mm and 73.5mm dimensions are known.

(1) Solve for the hypotenuse: C^2 = 19^2 + 73.5^2

Google: sqrt(19^2+73.5^2) = 75.916

(2) Solve for the angle: tan(x) = 19/73.5

Google: arctan(19/73.5) to degrees = 14.494

(3) The 50 degree dimension is known. Use this to find the angle of the bigger triangle.

50-14.494 = 35.506

(4) Solve for the distance to centerline: cos(35.506) = x / 75.916

Google: cos(35.506 degrees) * 75.916 = 61.7998

mhajicek · Oct 15, 2021

When I was in school my solution for this sort of thing was to write a program that could solve the general case, and test it with several known solutions.

PROBE · Oct 15, 2021

gregormarwick said:
extend the 40° line to where it intersects with the centre line by constructing a triangle from the 38mm flat.

You then are left with a single triangle that you can calculate the length you require.

Old, good trigo. Nothing compares to it.

CarbideBob · Oct 15, 2021

Is it me or did something change in this post?
The 19 or double comes from where and the 73.5 line did not look square to the world but it is now gone.
Wonder the most complicated way to solve or most math.
Sketching it in CAD is surely cheating.

Bobw · Oct 15, 2021

CarbideBob said:
Is it me or did something change in this post?
The 19 or double comes from where and the 73.5 line did not look square to the world but it is now gone.
Wonder the most complicated way to solve or most math.
Sketching it in CAD is surely cheating.

The 38 at the top.. And the 147 WAY over on the right.. Of course we all just
Ass-U-Me'ed that everything was symmetrical.

It looks a little wonky, pic of a monitor from a bit of a high angle.

The most complicated way?? I'm sure you could create enough geometry off of that to
make the math take you the rest of your life.. You could put in a 3 point tangent circle,
and then come off of that to calculate your distance.. That would be complicated, and
a bit stupid.. But.. eventually.. it would get you where you need to go.

CarbideBob · Oct 15, 2021

Bobw said:
The 38 at the top.. And the 147 WAY over on the right.. Of course we all just
.

And now the print is back and the hand sketch gone. Once there were two. Then there was one and now the other.
Must be a brain failure.

MazatrolMatrix · Oct 16, 2021

Orange Vise said:
View attachment 332225

Look at the skinny triangle first. 19mm and 73.5mm dimensions are known.

(1) Solve for the hypotenuse: C^2 = 19^2 + 73.5^2

Google: sqrt(19^2+73.5^2) = 75.916

(2) Solve for the angle: tan(x) = 19/73.5

Google: arctan(19/73.5) to degrees = 14.494

(3) The 50 degree dimension is known. Use this to find the angle of the bigger triangle.

50-14.494 = 35.506

(4) Solve for the distance to centerline: cos(35.506) = x / 75.916

Google: cos(35.506 degrees) * 75.916 = 61.7998

Thanks for that

CarbideBob · Oct 16, 2021

Lets do it the easy way.
Point on the line is -19 x and 73.5 y.
If one had to measure it on B-port rotary table or surface plate you would twist it so the line straight.
This simple.
One sine and one cos is needed. No need to solve the other axis.
Triangles are great but sometimes easier to rotate the world around zero.

MazatrolMatrix · Oct 17, 2021

CarbideBob said:
Lets do it the easy way.
Point on the line is -19 x and 73.5 y.
If one had to measure it on B-port rotary table or surface plate you would twist it so the line straight.
This simple.
One sine and one cos is needed. No need to solve the other axis.
Triangles are great but sometimes easier to rotate the world around zero.

In this case it was necessary for programming purposes.

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