Trigonometry Practice

Home Model Engine Machinist Forum

Help Support Home Model Engine Machinist Forum:

This site may earn a commission from merchant affiliate links, including eBay, Amazon, and others.

Rayanth

Well-Known Member
Joined
Jun 7, 2011
Messages
336
Reaction score
7
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 :
Trig1.jpg


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

- Ryan
 
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
 
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

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

Ken

dump.JPG
 
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’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” X 18tpi and 3/8” 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’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’s. Remember driver gears in the numerator, driven gears in the denominator.

Hope this helps someone who never knew
Regards, Ant
 
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
 
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
 
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
 
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


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

Ken
56688d1344178480-trigonometry-practice-dump.jpg

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.
 
....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 :
Trig1.jpg


.....- Ryan
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.
 
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

sphring.JPG
 

Latest posts

Back
Top