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Maths problem just for fun

Get that old nut of yours warmed up with this one.

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Buffer16/12/2021 11:31:07
343 forum posts
155 photos

I found this in a book yesterday and to be honest I can't work it out. I know the answer as I have a good CAD program. But what about you, can you do it the old fashioned way?

How high above the surface of the part does the ball protrude?

20211216_112250.jpg

RobCox16/12/2021 11:57:41
59 forum posts
20 photos

99.4 thou if my maths is correct.

If the slot width is 2W and the ball diameter = 2R and the angle of the slot sides to the vertical = A:

The ball contacts the sides a height of R.sinA below the ball centre.

The distance across the slot at the contact points is 2R.cosA, so the height below the top of the slot is

(W-R.cosA) / tanA, giving the height of the top of the ball above the top of the slot as

H = R + R.sinA - (W-R.cosA) / tanA

For R = 0.5, W = 0.625 and A = 90-75 = 15, that gives the answer above.

Shoot me down if I'm wrong.

Rob

JasonB16/12/2021 12:09:22
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23042 forum posts
2769 photos
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0.414"

My method

maths.jpg

1. Work out how far below ctr of ball it makes contact 0.129 and also the horizontal distance 0.966

2. Work out how far down the taper the 0.966 width is 0.215

3 height is the 0.129 plus radius of 0.500 less the 0.215.

Answer or at least my answer is 0.414

Edited By JasonB on 16/12/2021 12:14:16

RobCox16/12/2021 12:20:36
59 forum posts
20 photos

My answer is clearly wrong, because half of 1.125 is not 0.625embarrassed

Edited By RobCox on 16/12/2021 12:20:56

JasonB16/12/2021 12:25:22
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23042 forum posts
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So is mine now I have done it in CAD, method was right but I got the distance down that the points of contact are wrong as I used TAN not SIN, won't post the right answer so others can have a go

Edited By JasonB on 16/12/2021 12:28:52

RobCox16/12/2021 12:32:06
59 forum posts
20 photos

Jason, I agree with your method, but I get the height of the contact below the top (x in your diagram) to be 0.0795/tan15 = 0.2968, which gives me a revised answer of 0.3326.

Thank goodness I haven't got to worry about maths exams any more!

JasonB16/12/2021 12:55:36
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That's it.

Nextdevil

not done it yet16/12/2021 13:07:32
6887 forum posts
20 photos

Nothing other than a cursory glance, but I think it needs a couple of simultaneous equations to sort out the contact point. Might look at it later…

Martin Connelly16/12/2021 13:08:11
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2180 forum posts
227 photos

This is my take on the solution. It requires solving the equation to find angle alpha using trig identities which I did not include on the page as it is messy and seems a bit unnecessary to show here. The point is it gets the same result a different way from those above.

p1160224 (2).jpg

Martin C

Jon Lawes16/12/2021 15:31:09
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984 forum posts

I can't help but feel inferior to the brains knocking around on this forum. I'd almost certainly have had to google the technique; I am for some reason unable to retain mathematics techniques and such.

Well done to the examiner and to those who attempted an answer!

Buffer16/12/2021 17:01:28
343 forum posts
155 photos

Well as Jason asked for it, here it is. I take no credit for this and I can't do it without CAD.

If the ladder just touches the cube how far up the wall does the ladder touch the wall? Good luck you'll need it.

20211216_165257.jpg

Brian Wood16/12/2021 18:06:54
2579 forum posts
39 photos

I cheated and solved it by construction!

The answer there is 9.625 feet

Brian

JA16/12/2021 18:25:50
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1401 forum posts
81 photos

Agreed.

Wedge sum

I have met Buffer's problem before and it was deemed un-solvable. I have just written the equations down: three simulanteous quadratics. My evening meal calls. I am sure someone will solve it by the time I next look at the website.

JA

Bob Mc16/12/2021 18:26:57
231 forum posts
48 photos

Ladder box wall problem...

It can be done, I spent many a happy hour trying to solve this one some years ago, and although I didn't find a solution myself the problem can be solved, here is one solution.

**LINK**

Pete Rimmer16/12/2021 18:47:24
1256 forum posts
69 photos

I coudn't do the box-and-ladder one but for sure it has two possible solutions. Only the visual representation (and the fact that no-one would use such a low-angle ladder) suggests that the base of the triangle is less than the height. Mathematically there are two values of X where the line meets the cube.

JA16/12/2021 20:07:50
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1401 forum posts
81 photos

I have just looked at the link. It is the same two solutions turned through ninety degrees.

It would be interesting see if complex numbers would make the solving easier. Too old for that now.

JA

DMB16/12/2021 21:09:08
1350 forum posts
1 photos

With the first diagram, I didn't actually try maths but thought "ball-park" - the height of a vertical radius above the flat top is approximately 2/3 of the total radius. Total radius is 1/2", (16/32) so 2/3 is about 10, or 5/16, so it seems my guess was quite close!

pgk pgk16/12/2021 21:40:47
2603 forum posts
293 photos

Surely the first problem has a simple approach.
A line drawn vertically up from the ball/wall contact point meets a line extended from the block surface to give an 'inverted' triangle. Base of that is 1/2 (width of VGroove minus DiameterBall). Tan(15deg) then gives the inverted height to subtract from ball radius?

The second problem I’m struggling with. I think the solution rests somewhere in the fact that the hypotenuse of the triangle above the cube is also X such that there are simultaneous equations with Cos(Theta)=x/10=(x-2)/x. Also Tan(Theta)=(x-2)/2 and Sin(Theta)=2/X. I guess one needs to be familiar with the Trig reciprocals, which gets beyond the math I learned.?

pgk

duncan webster16/12/2021 22:17:37
4119 forum posts
66 photos

My answer to problem 2 is 9.677 ft, but I had to use a Bisection routine to get to it, otherwise I think you have to solve a 4th power equation.

I've checked my answer on CAD

RobCox16/12/2021 22:31:58
59 forum posts
20 photos

I could try and pretend I'm really whizzy with maths by giving a solution, but I'm not, so I'll instead refer you to a youtube channel I've watched from time to time. The channel creator's name is Presh Talwalker and he solves this exact problem in his video " the ladder and box problem - a classic problem".

Me, I just got bogged down in quadratic equations that looked like they might turn into quartics...

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