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2 Ball Bearings in a Tapering pipe

Hi,

I'm new here and after any help that can be given. I've finally decided to throw in the towel on a few questions and ask for help on them. This question is from a revision sheet handed out in class. My exam is on the 2nd (Sweating)

Attachment 22113

Any help is extremely appreciated.

Thank you.

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Re: 2 Ball Bearings in a Tapering pipe

Quote:

Originally Posted by

**Dcoz** Hi,

I'm new here and after any help that can be given. I've finally decided to throw in the towel on a few questions and ask for help on them. This question is from a revision sheet handed out in class. My exam is on the 2nd (Sweating)

Attachment 22113
Any help is extremely appreciated.

Thank you.

1. I've modified your drawing a little bit: The greyed triangle is a **right **triangle.

2. The distance between the midpoints is 45.97. The difference of the 2 radii is 6.

The smallest angle in the right triangle equals $\displaystyle \frac12 \theta$.

Calculate the value of $\displaystyle \theta$. You should come out with 15°.

3. Use proportions, the Sine and Cosine function to determine the missing lengthes.

Re: 2 Ball Bearings in a Tapering pipe

Thank you. I completely understand how you got to the angle using your clear instructions, but unfortunately I'm still strangling with finding the internal diameter. My assumption is I would have to determine the two equal sides of the triangle and although I am somewhat familiar with proportionality, I'm lost on how to use it to calculate the rest.

Again thank you for the kind help so far!

Re: 2 Ball Bearings in a Tapering pipe

Ok clearly I have to find the total height, but still no idea where to begin with proportionality.

Re: 2 Ball Bearings in a Tapering pipe

Ok, is it correct if I used the radius of the larger bearing as 15mm and opposite angle (of the tapered pipe) as 7.5 degrees in a right angled triangle making the height 113.93 and simply adding the extra 15 and 4.34mm to it?

Re: 2 Ball Bearings in a Tapering pipe

Surely the internal diameter of the tapered journal at the two points the bearings are located must be the internal diameter of the journal at those two points, which must be the diameter of the bearings, then using ratios can you not find the answer?

Re: 2 Ball Bearings in a Tapering pipe

I've tried to figure it out, but I don't have enough practice with ratios.

Re: 2 Ball Bearings in a Tapering pipe

I'm able to calculate the ratio between the two bearings 5:3, but stuck trying to factor in the 4.34mm into it. I'm sure I'm gonna kick myself when it's figured out.

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Re: 2 Ball Bearings in a Tapering pipe

Quote:

Originally Posted by

**Dcoz** Ok, is it correct if I used the radius of the larger bearing as 15mm and opposite angle (of the tapered pipe) as 7.5 degrees in a right angled triangle making the height 113.93 and simply adding the extra 15 and 4.34mm to it?

Here is how I would have done this question:

1. The radius of 9 mm (smaller sphere) increases to the radius of the larger sphere by adding 6 mm to the radius on the way of 45.97 mm. If this way runs up to the upper rim of the pipe then the length x would be added to the 9 mm:

$\displaystyle \dfrac x{65.31}= \dfrac6{45.97}~\implies~x=8.524255$

2. The green painted triangle is a right triangle. The smaller angle of the triangle is $\displaystyle \frac12\ \theta = 7.5^\circ$. One leg of this angle has the length

$\displaystyle l = 9 + 8.524255\ mm = 17.524255\ mm$

3. Half of the diameter D is the hypotenuse of the right triangle. Use the Cosine function to calculate the hypotenuse H:

$\displaystyle \frac12\ D = H = \frac{17.524255}{\cos(7.5^\circ)} = 17.6755$

Therefore D = 35.35 mm

Re: 2 Ball Bearings in a Tapering pipe

Thank you for your method which appears to be more accurate than what I had attempted. I obtained a diameter of 35.086, if you have the time, do you know why your measurement of the diameter is ever so slightly larger or is the difference nothing to worry about? Also, I'm trying to understand why you may have drawn the radii at a slight angle rather than horizontally? And lastly, are you able to explain why the green painted triangle's smallest angle is equal to the original angle theta we were trying to find. Thank you so much for you time!

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Re: 2 Ball Bearings in a Tapering pipe

Quote:

Originally Posted by

**Dcoz** Thank you for your method which appears to be more accurate than what I had attempted. I obtained a diameter of 35.086, if you have the time, do you know why your measurement of the diameter is ever so slightly larger or is the difference nothing to worry about?

1. The measures of the lengthes are given by an accuracy of $\displaystyle \tfrac1{100} \ mm$. So (in my opinion) a deviation of nearly $\displaystyle \tfrac{30}{100}\ mm$ is intolerable.

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Also, I'm trying to understand why you may have drawn the radii at a slight angle rather than horizontally?

2. The inner surface of the pipe is tangent to the sphere. The radii of the sphere form a right angle with the inner surface of the pipe in the tangent points (The tangent points are located on a circle which is placed on the surface of the sphere). Since the surface of the pipe is "tilted" by an angle of $\displaystyle \tfrac12\ \theta$ the radius has to be tilted by exactly the same angle.

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And lastly, are you able to explain why the green painted triangle's smallest angle is equal to the original angle **half theta** we were trying to find. Thank you so much for you time!

3. Have a look at the sketch. You have 2 **right **triangles with the common angle $\displaystyle \alpha$. That means that the remaining angle, I've labelled it $\displaystyle \tau$, must be equal.

Re: 2 Ball Bearings in a Tapering pipe

I am unable to thank you enough for all your help. I do enjoy this subject, but when you hit a wall it's very difficult to not want to turn away from it. Once you've reached that "ah ha" moment, you forget instantly the pain you've long suffered. Once again I thank you for that!