# Trigonometry Practice



## Rayanth (Aug 5, 2012)

As I go through Algebra/Trig ("Pre-Calculus") at school, I am stumbling on a number of sample problems that could come in useful for our hobby. I have been re-learning all this stuff under a new light, trying to adapt what I'm learning to what we do in engine design and building.

Would anyone be interested in me posting some of the problems, to see if they could work them out themselves? I will admit it's not the exactly the easiest mathematics, but it's certainly interesting and handy in some cases.

For example, one I just solved (correctly, on my own, it was not assigned as homework) determined the area of the section in between three cotangent circles. Not insanely useful, but an interesting exercise that might have applications down the road.

A much simpler example could be: Let us say that you have a piece of round bar, of diameter 1.5 inches. You require a piece using that outer radial section, but with a flat side of width 1.25 inches. How far down from the opposite edge do you need to make the cut? (or from center)
Solve for x and/or y :






If there is interest in this, I can post more complex problems, to get the old gears lubricated and ticking again =)

- Ryan


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## Tin Falcon (Aug 5, 2012)

Ryan:
I am glad you are enjoying learning the math. Math is certainly foundational to machining.  the problem you show and many others like it is in the math section of the machinery handbook and covered in a little more detail in the guide book.
You may want to pick up a copy of _Mathematics at Work _from Indusrial  Press. 
covers mostly the same material but in more detail . 
I forced my son to do this very calculation when he was 14 and building a steam engine.
Tin


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## Rayanth (Aug 5, 2012)

Tin Falcon said:


> Ryan:
> I am glad you are enjoying learning the math. Math is certainly foundational to machining.  the problem you show and many others like it is in the math section of the machinery handbook and covered in a little more detail in the guide book.
> You may want to pick up a copy of _Mathematics at Work _from Indusrial  Press.
> covers mostly the same material but in more detail .
> ...



Tin, I'm not sure if I was properly understood there =) I have the knowledge to solve the above problem, I just thought I might supply a few problems for others to try their hands at, if they so wished...

- Ryan


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## Tin Falcon (Aug 5, 2012)

Ryan : not implying you do not know how to solve it . If that was the case I would have posted the answer and the formula. 

You want to share you math skills and experience that' s fine. I was simply posting resources for machine shop math so folks can find the formulas and math tools they need for shop problems. 
Off the top of my head that problem would be unsolvable. I do not do them every day these days. but a look at the formula and a minute or two of plug and chug I would have the height from the cord to the circular segment then simple math to find the distance x and y. 
Tin


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## Hopefuldave (Aug 5, 2012)

Easy trigonometry, how I miss it - a bit of squaw on the hippopotamus and the squaws on the other two hides 

If only I could do the square root of 11/64ths in my head...

Dave H. (the other one)


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## Antman (Aug 5, 2012)

Sine rule is very neat.  If you have 3 points not in a straight line you have a triangle if you join the 3 points with 3 straight lines.  Call the sides a, b and c and angle A is opposite side a,  angle B opposite side b etc.
   Then
    a/sinA  =  b/sinB  =  c/sinC   and that is also equal to the radius of the unique circle that passes through each point.  Useful for measuring bolt holes and the like.  You might have to use the cosine rule to get the value of the angles.  Your 3 buck calculator gives the trig ratios to 10 places,  so your only inaccuracies come in your measuring of straight lines.
 Ant


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## kuhncw (Aug 5, 2012)

Good idea, Ryan.  I think posting a few problems to stir the cobwebs is a good idea.

The recent discussion on flywheel inertia, in John's thread on his nice V Twin, had me getting out the books to dig back into calculating moment of inertia.

Regards,

Chuck


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## Ken I (Aug 5, 2012)

Assuming 0.75 is the radius, then from Pythagarus

0.75^2 = x^2 + 0.625^2

therefore x = ((0.75^2)-(0.625^2))^0.5 = 0.4245

Ken


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## Tin Falcon (Aug 5, 2012)

Ryan: 
I hope you do not feel I am discouraging your endeavor. Math is not not a hot  topic like politics but people do have there feelings about it. some are very comfortable with it others avoid it. calculus and deferential equations are words that strike fear in mighty men.  
I am one that struggles with math. I feel better about it when I see the practical applications and not just dreary repetitive exercises.  
I enjoyed Algebra in college a kamakazzi summer course 4 nights a week for four weeks. one could not put off the home work. And I was working full time as well. 
The teacher was a cute freckle faced slender  Millbilly Girl. An excellent teacher.
Trig was fun also I actually like vector addition.  The teacher was an adjunct who day job was HS math. so he knew the material but did not know how to treat adults like adults. Calculus a lot of work. 

I have found Industrial Press Books to present math in a simple practical fashion .The presentation is concise and relevant to machining. 

So  Ryan spread some of that youthful enthusiasm to the old  guys here. 
Tin


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## Ken I (Aug 5, 2012)

Try this - can be solved by mental arithmetic - but the sine rule works if you must.

Ken


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## Cogsy (Aug 5, 2012)

Ken I said:


> Try this - can be solved by mental arithmetic - but the sine rule works if you must.
> 
> Ken


 
Math class was a long time ago, and I'm struggling with either 0 or 360...

Al.


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## Antman (Aug 5, 2012)

Judging by the number of people on various forums that want a 127 tooth gear to cut imperial threads on metric lathes or vice versa, here&#8217;s a little arithmetic relationship that everyone who uses a simple lathe, where you have to remove and swap cogs manually, can use, but not everyone knows.

1mm X 40/42 X 8/3 = 0.99987501 inches i.e. a tenth of an inch to better than 99.9875% numerical accuracy. 

For example in my case I have a lathe with a 2mm pitch leadscrew. I like to use 3/8&#8221; X 18tpi and 3/8&#8221; X 24tpi nuts and bolts in some of my lathe tools as they are surprisingly available in metric South Africa and they have nice 14mm hexes across flats. All the M8 bolts in my lathe have 14mm heads (most M8&#8217;s have 13mm heads) so a 14mm spanner is always to hand. My lathe spindle has a 40T gear. 

So 18tpi is 
2mm X 40/42 X 50/60 X 80/90

and 24tpi is
2mm X 40/42 X 50/75 or 40/60 

And similarly for other tpi&#8217;s. Remember driver gears in the numerator, driven gears in the denominator.

Hope this helps someone who never knew
Regards, Ant


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## Tin Falcon (Aug 5, 2012)

> Try this - can be solved by mental arithmetic - but the sine rule works if you must.


OOH ken that is being mean . I know you did not draw that to scale ;D
Tin


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## Ken I (Aug 5, 2012)

Tin Falcon said:


> OOH ken that is being mean . I know you did not draw that to scale ;D
> Tin


 OOh - you are so right - it caught me out when I first saw it - I eventually solved it using the sine rule - and then mentally kicked myself.

Ken


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## Tin Falcon (Aug 5, 2012)

I looked at the thing I saw that 4+3 =7 something did not seem right. so I started to construct the triangle . that is when I realized it was not a triangle at all. a good one indeed.
Tin


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## Rayanth (Aug 5, 2012)

Tin Falcon said:


> Ryan:
> I hope you do not feel I am discouraging your endeavor. Math is not not a hot  topic like politics but people do have there feelings about it. some are very comfortable with it others avoid it. calculus and deferential equations are words that strike fear in mighty men.
> I am one that struggles with math. I feel better about it when I see the practical applications and not just dreary repetitive exercises.
> I enjoyed Algebra in college a kamakazzi summer course 4 nights a week for four weeks. one could not put off the home work. And I was working full time as well.
> ...




Tin, I was not discouraged, merely misread the tone of your post. It seems a few people at least may be interested, so I will wait a few days and post the answer, then another problem.

- Ryan


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## Tin Falcon (Aug 5, 2012)

There are many perspectives here . With math often many approaches  to obtain a solution.the same thing with machining many approaches to solve a problem or make a part. 
Tin


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## JohnT (Aug 6, 2012)

Ken I said:


> Try this - can be solved by mental arithmetic - but the sine rule works if you must.
> 
> Ken







I have a bottle in front of me (Smithwick's, which I'm enjoying thoroughly; several actually) so I might be a wee bit off base here.
Actually, I would use the Law of Cosines, rather than the Law of Sines.  


Wait a minute. Looks like I spoke too soon... I kept pondering Kens statement "...can be solved by mental arithmetic.".... I see what Ken means now: 16+49 - 9 = 56. divide that by 2*4*7; Which is 56/56 = cos(1) Therefore angle x = 0 and there is no solution. HaHaHaHa! That's great!! I had forgotten from Trig, when dealing with cases of non right triangles, first make sure  by inspection that the sum of the two shorter legs is GREATER than the longest side. I'm not accustomed to looking at triangles with no solutions, since that never happens in the machine shop. Nevertheless, an amusing little problem.


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## JohnT (Aug 6, 2012)

Rayanth said:


> ....Let us say that you have a piece of round bar, of diameter 1.5 inches. You require a piece using that outer radial section, but with a flat side of width 1.25 inches. How far down from the opposite edge do you need to make the cut? (or from center)
> Solve for x and/or y :
> 
> 
> ...


 An interesting variant of this problem is when the radius of the circle is NOT given; but you are given x and y: solve for the radius.
 It has come up but rarely in my professional career as a CNC machinist and at home as well, but nevertheless it has turned up. Matter of fact, coincidently a fellow machinist asked me how to solve this very same type of problem just two weeks ago, as it turns out! Somewhat unusual.


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## Ken I (Aug 6, 2012)

O.K. here's a difficult one - its not really a trick question like my earlier triangle and you can work out the answer the hard way (all the formulae are given).
There is a simple method if you apply your mind.
(I did it the hard way before I figured it out).

It would appear from first glance that there is insufficient information to answer the question - but there is.

Ken


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## AlanHaisley (Aug 16, 2012)

Ken I said:


> O.K. here's a difficult one - its not really a trick question like my earlier triangle and you can work out the answer the hard way (all the formulae are given).
> There is a simple method if you apply your mind.
> (I did it the hard way before I figured it out).
> 
> ...


Ken,

Nice problem! When I started on this, it seemed to me that - as you say - I needed at least one more piece of information.

Finally, I decided to see what the answer would be if I just tried a couple of sizes of holes; I'm afraid that I was still a bit muzzy minded at the time. 

For one hole I picked, of course, the largest possible sphere. Obviously this gave me a drill diameter of zero. It also gave me a sphere diameter of 100mm.

For the other hole I picked one with a diameter the same as the depth of hole. Interestingly, I got the same answer for the volume. Surprise, surprise!

Finally I realized that I could indeed express D, d, and h in terms of L.

I decided not to post my work here in case others are playing around with this; still, my hints should lead others to the solution.

Even having worked this out, the result is counter-intuitive.


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## Ken I (Aug 16, 2012)

Counter intuitive indeed, the answer is a constant - hardly credible really - glad you liked it.

I posted it before some time back but no one "bit"

Ken


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## JohnT (Aug 18, 2012)

Given a circle of radius R with a chord with a length of 26in. At the midpoint of the chord there is a perpendicular bisector that intersects the circle and has a length of 3.75in. Find R.
This is a real world problem that came up once or twice in my career. It can be solved by 3 simultaneous equations in 3 unknowns or reduced row echelon form of a matrix , but there a couple of ways to solve it directly far faster.


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## Maryak (Aug 18, 2012)

R = (c^2+4h^2)/8h

R = (676+56.25)/30

R = 24.4083333


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## Ken I (Aug 19, 2012)

I got the same answer as Bob from Pythagoras - ?

Ken


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