**Pie-eyed piper Problem 206** — as this month’s problem by Andrew R. Hogg of Mason, Mich., demonstrates.

“Why, I’ve solved our dilemma!” quipped Lucius Bluff. “We’ll just use three pieces of I-beam to support it. I’ve already had them cut to fit.”

The project was to support a pipe with a 56.25-in. outside radius inside a larger pipe with a 120-in. inside radius. The centers of both pipes must be identical.

Bluff had cut the I-beams such that each is tangent to the inner pipe and their inside edges meet at the inside of the outer pipe, forming an isosceles triangle. (See figure.)

Are the centers identical with such an arrangement? If not, assume the center of the smaller pipe is below the center of the larger pipe. How far below the center of the outer pipe is the center of the inner pipe?

Send your answer to:

Fun With Fundamentals

POWER TRANSMISSION DESIGN

1100 Superior Ave.

Cleveland, OH 44114-2543

Deadline is July 10. Good luck!

*Technical consultant, Jack Couillard, Menasha, Wis.*

**Solution to last month’s problem 205** — You know your limits if you answered yes. Here’s where things left off:

First find the initial velocity of the lure. The distance it travels is:

The distance divided by the time, 0.167 sec, is 62.8 fps

Next find the release angle. Each 5 min. of the clock is 30 deg. At 10:00 the angle from the vertical is 60 deg, and this is the release angle of the lure.

From physics we know that

Where

*v _{o}* = Initial velocity, fps

*a*= Release angle, deg

*t*= Time to reach zenith, sec

*= Gravitational constant, 32.2 ft/sec*

g

g

^{2}Let s be the distance the lure travels. Also from physics for curvilinear motion,

*s* = *v _{a}* cos

*a*x 2

*t*

Substituting for *t* in (1),

But we have to remember that at 10:00 the lure is (6.66 ft)(cos 30) 55.8 ft behind the railing. So the distance traveled from the railing to the fence is 100.2 ft. The dog got it!