May 1, 2005
Best Buy sympathies I also became entrapped by the Best Buy Corporate Policy as it relates to purchases and the return/replace/repair customer agreement.

Best Buy sympathies

I also became entrapped by the “Best Buy Corporate Policy” as it relates to purchases and the return/replace/repair customer agreement. But instead of continuing my debate with store managers and employees, who could only submit they were held to the Best Buy Policy, in addition to the fact that it was clear in the fine print that the store could not be held accountable, I went directly to the manufacturer, Sony. What I found is that Best Buy has an agreement with Sony, in contradiction with Best Buy Policy, requiring them to assume a more gracious protocol with customers on Sony's behalf. Failure to do so puts them at odds with their contract and exposes them at the very least to losing their status as an approved distributor of Sony products. In the end, I found complete satisfaction to the dismay of Best Buy employees who couldn't believe their customer “unfriendly” policy couldn't be enforced. You may want to contact Sony, because I found them to be completely reasonable and fair. They supplied me with contact people and numbers to effectively close my successful debate with Best Buy.
Rob Niemiec
V.P. of Manufacturing
Motor Specialty Inc.

I feel your pain. What a lousy experience. One has to believe that yours is not an isolated case. When I experienced something similar with a digital phone service, adding bells and whistles to my plan that I did not order, I referred it to the State Attorney's office. It got results, not fast, but eventually. Turns out others had also called and written in, which resulted in a class action suit. You might want to do the same.
Dorian Olivera
West Hartford, Conn.

With regard to your “Best Buy Blues” editorial in the March 2005 issue, be advised that I share your horror at Best Buy's merchandising tactics. After returning four separate defective purchases in a row, I voted with my feet. Neither I, nor my family, have been back to a Best Buy store in two years. It is a wonder that I still see cars in their parking lots. Hope apparently springs eternal.
William Woodcock
Lockheed Martin Space Systems Co.
Civil Space Programs

More fun with FWF

I'm glad to see that the “Fun with Fundamentals” has made it back into Motion System Design. I always enjoy the problems; keep them coming.
Paul J. Carter, PE
Mechanical Engineer
Closet Maid

I have seen firsthand the damage that a birdstrike can do to an airplane. I spent 15 years in the aviation industry, and was involved with birdstrike projects as part of aircraft and equipment design. The results are not pretty.

Thanks for bringing back “Fun With Fundamentals.” I was introduced to the column by a “senior” engineer back in 1993. And now I challenge my “junior” engineers with it. Thanks again for publishing this great column for engineers.
James V. Zappa
Engineering Services Group Leader, Hardlines Division
Bureau Veritas Consumer Products Services
Amherst, N.Y.

Thanks for bringing back the fun. It's been a long 15 months without “Fun with Fundamentals.”
Tony Bullard
Bullard Welding Co.
Chelsea, Vt.

I'm delighted to see “Fun with Fundamentals” is back. Thanks.
Roy Carlson

Alternative FWF solution

Problem 275 is reminiscent of an example from a textbook in classical mechanics that I studied over forty years ago (how time flies). The parametric equations for the x and y coordinates of the projectile in terms of time t are x = v t cos a and y = v t sin a + g t2/2 + 8, where v = velocity = 500 ft/sec, a = 15°, and g = gravitational acceleration = 32.17 ft/sec2.

The height of the projectile at a distance of 75 ft from its launcher is determined by calculating the time required for the projectile to travel a horizontal distance of 75 ft using the first equation above and then using that time to calculate the height of the projectile at that time using the second equation.

The time to travel 75 ft in the horizontal direction is t = 75/(500 sin 15°) = 0.1553 sec, and the height at this time is 27.71 ft. Thus, the projectile passes safely over the plane.

The time required for the projectile to reach the ground is calculated by setting y = 0 in the second equation and solving the quadratic for t. Using the quadratic equation (and my trusty HP 33s calculator) the time for the projectile to reach the ground is 8.1067 sec. The horizontal distance that the projectile travels during this time is calculated from the first equation - x = 500 * 8.1067 * sin 15° = 3,915.24 ft.
Allan H. Reed
Senior Research Electrochemist
Technic Inc.

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