Motion System Design
Fun with Fundamentals: Problem 162

Fun with Fundamentals: Problem 162

Intolerable circumstances and Solution to Problem 161. What you see is not always what you get, as this month’s problem by Kevin Kuester demonstrates

Problem 162 — What you see is not always what you get, as this month’s problem by Kevin Kuester demonstrates.

A volley of unprintable language issued forth from the electrical contractor’s trailer on the site of Phineas Gotrocks’ new summer home on Lake Hyhrent. His villa was to be the last word in luxury, and Gotrocks and the electrical contractor, Nicholas Fritz, were each trying to have that last word.

“You want stereo speakers in the bathroom and in all 10 bedrooms; three-way lighting in all seven bathrooms, floodlighting for the tennis courts and heated swimming pool, plus little lampposts in the gardens and driveway? It can’t be done,” shouted Fritz.

“Nonsense, my good man,” drawled Gotrocks as he pointed to the proposed electrical diagram. “I know something of electrical work, and all you have to do is run three single-strand wires through these pipes here instead of two.”

“They’ll never fit!”

“Certainly they’ll fit! Why the pipes have 0.500-in. inside diameters — with a tolerance of 60.010 in.”

The wires each have circular cross sections and outside diameters of 0.250 in. 60.005 in. If there must be a 0.020-in. radial difference between the pipe ID and a circumscribing circle around the three wires (to allow easy assembly), will the pipes be able to accommodate three wires?

Send your answer to:

Fun with Fundamentals POWER TRANSMISSION DESIGN Penton Publishing 1100 Superior Ave. Cleveland, OH 44114-2543 Technical consultant: Jack Couillard, Menasha, Wis.

Solution to last month’s problem 161 — You know a load of waste material when you see one, if you answered 5.94 years. Here’s how Wurme piled it higher and deeper:

First convert the yearly production rates and FC capacity from weight to volume measures:

For FC production rate:

For BA production rate:

For capacity of FC in top layer:

From Wurme’s data we know that there is 1,363,000 yd3 of capacity for the base layer (1,750,000 yd3 total capacity 2387,000 yd3 top layer capacity.)

Let: A = Production rate of BA, given as 37,037 yd3/year B = Production rate of FC, given as 243,810 yd3/year N = Number of years capacity, years x = Portion of FC that goes to base layer, no units

Since the base layer is comprised of both FC and BA, we can set up the following equation for capacity:

The remainder of the FC goes to the top layer. We know that the capacity for FC in the top layer is 301,479 yd3, so we can set up the following equation:

Substitute the value of N in Equation (2) for N in Equation 1. Cross multiply to get rid of the denominators:

1,363,000B - 1,363,000Bx = 301,479A + 301,479Bx

3.32 x 1011 - 1.12 x 1010 = 4.06 x 1011x

Plug in the values for A and B and solve for x:

Plug the values for x , A, and B into (1) and solve for N.

[37,037 yd3/year + (0.79)(243,810 yd3/year)]N = 1,363,000 yd3

N = 5.94 years

Bluff will have a tough act to follow when he takes over Wurme’s job as vice president of quality control, Mr. Wurme being unavailable for some years to come.

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