My calculator is overheated from the last two BTCs, so this time let’s do a little trivia. In my younger days I had occasion to visit one of my Arkansas cousins who was trying to make a killing selling surplus military equipment. You younger folks may not know, but it was a common practice in those days for the military to sell expensive electronic parts and equipment for pennies per pound. It was amazing the good deals you could find. Well my cousin Otis, had just scored 317 pounds of little glass bulbs with five wires protruding radially around the outside of each one. The thing looked like it could have been the hub and spokes of a very tiny spinning wheel.
Anyway, Otis wanted me to help write the ad for the ham swap meet that was coming up and wasn’t sure how to describe his bounty. I gave him a hint that these vacuum tubes were named for the fruit/nut of a common tree. With just this much information, can you tell me what the common name for this type of tube was? And to really impress me, what were the five leads connected to inside the envelope?
Reply to Butch Shadwell by January 26 at firstname.lastname@example.org (email), 904-223-4510 (fax), 904-223-4465 (v), 3308 Queen Palm Dr., Jacksonville, FL 32250-2328. (http://www.shadtechserv.com) The names of correct respondents may be mentioned in the solution column. The solution to this challenge will be published in the February Region 3 Newsletter, which will be available in February (surprise!) at http://ewh.ieee.org/reg/3/enewsletter
The BTC last issue was about interplanetary travel, in a very simplified form. It had to do with the fact that Mars is in perigee and what if visiting Martians were traveling in a space craft as described. “It is a mere 30 million miles away…accelerate your craft at a constant rate of 9.8 meters per second squared, how long will it take to arrive and stop at Mars orbit? Fortunately, you can swing your ship around so that the thruster can be used to slow as well as speed up your ship. For extra credit, tell the highest speed you ship will reach on this trip. Let’s ignore the fact that the target is actually moving during this flight and any gravitational effects. If you’re really feeling intelligent today, tell me the total amount of energy used for the trip in kilowatt-hours if our craft has a mass of 5000Kg.”
This one takes more number crunching than most, so here we go. First the acceleration curve is only half the trip long, so I convert 15E6 miles into 24.1E9 meters. Then we know that X=1/2(At^2), so t=(2X/A)^1/2. We get t=70.1E3 seconds for half the journey. This means that the whole trip takes about 39 hours. In case you didn’t catch it, 9.8 m/s^2 also happens to be the rate at which the pavement accelerates toward your face when you trip over the chalk line in a field sobriety test. Then V=At, since the highest speed is at the midpoint we get Vmax=687E3 m/s or 1.54E6 mph (c/435). Note that we are also ignoring the relativistic change in mass of our space ship. So for the last part you should know that 1 watt-second = 1 joule = 1 Newton-meter. Our ship’s engine is exerting a constant force of 49E3 Newtons for the entire trip (F=ma). Times the distance of 48.2E9 meters, we get 2.36E15 watt-seconds. After one more conversion we get 656E6 KWH, but I bet you already knew that. – Live long and prosper.