High School Math-Physics SMILE Meeting
27 February 2001
Notes Prepared by Porter Johnson

Bill Colson (Morgan Park HS, Math)
touted and passed around the book Rube Goldberg Inventions by Maynard Frank Wolfe [Simon & Schuster 2000] ISBN 0-684-86779-9. There was a recent article about a Rube Goldberg Machine Contest in the 26 February 2001 Chicago Tribunehttp://www.chicagotribune.com/.

You could almost hear the wheels turning as the small group of high school students huddled and talked and tinkered with their project. But that was the problem -- the wheels weren't turning -- and their entry in the Rube Goldberg Machine Contest wasn't ...

Fred Schaal (Lane Tech Park HS, Math)
made a presentation on finding the lateral surface area of pyramids with a regular polygonal base. He illustrated the point with a pyramid with a square base. The lateral surface consists of four equivalent triangles, of base b and height s.  The area of each triangle is ½ b s, where the height s is the slant height of the pyramid, and the lateral surface area of the pyramid is

Lateral Area = 4 ( ½ b s) =  ½ (4 b s).
This can be expressed in terms of the base perimeter p = 4 b and the slant height s as
Lateral Area = ½ p s.
For an n sided regular polygonal base of base length b and slant height s, the perimeter is p = n b and the lateral surface area is
Lateral Area =   ½ n b s =  ½ p s.
By taking the limit as the number of sides becomes infinite, one obtains the result
Lateral Area =   ½ n b s =  ½ p s.
for a (right circular) cone, where the base perimeter p  is the circumference of a circle of radius r, or p = 2 p r. Thus we obtain the standard expression for the lateral surface area of a cone,
Lateral Area = ½ p s = ½ 2 p r s = p r s .
You can also calculate the lateral surface area by rolling the cone about its apex, and computing the area of the sector so obtained, corresponding to opening angle q = 2 p r / s and radius s. The result is
Lateral Area = ½ s2q = ½ s2 (2 p r / s) = p r s
Very good, Fred. [Fred also reported a gigantic crane being installed on a construction site at the corner of Western Avenue and Lake Street.]

Earnest Garrison (Jones Academic Magnet HS, Physics)
addressed the problem of teaching Electricity and Magnetism to the Nintendo Generation. He first showed us a coil that produced a high voltage spark when touched to metal objects.  He recommended using a Basic Electronic Component Kit produced by Elenco Electronics Inc http://www.elenco.com/ [along with a manual of experiments]:

Kit Model PK-101
Omnitron Electronics
600 S. Military Trail
Deerfield Beach Fl 33442
1 - 800 - 379-6664
The kit, which transforms any standard breadboard into an Electronics Learning Center, contains the following items:
transistor, diode, led,
capacitors, resistors, potentiometer,
wires, circuit board.
Earnest gave us handouts on two experiments:
#1: The Light Bulb
#2: The Brightness control
He used a digital multi-meter, and a crank generator with the kit as accessories. He also described hooking the battery to a crank generator, which causes it to run backwards. He felt it was important to bring current "chip technology" into the classroom. He described a conducting material as being like Cheerios™ in milk, and an insulator as being like dry Cheerios™ . He described eddy currents set up in a copper tube, that caused it to fall slowly through a region of strong magnetic field, to illustrate that electric currents produce magnetic fields, and vice versa. Good, Earnest!

Bill Shanks (Joliet Central retired; Joliet Junior College Music Student)
took advantage of the after-Christmas sales to purchase a string of 70 Christmas lights for $1.50, reduced from $5.99. By removing bulbs he was able to determine that there were 2 strings of 35 bulbs hooked in series. The current through the bulbs was 120 mA, and the rms Voltage across each bulb was 120 Volts/ 35 = 3.5 Volts. Thus, the internal resistance of each bulb would be r = 3.5 Volts / 0.120 Amperes = 30 Ohms.

Bill next pulled out a light emitting diode (LED), which required a Voltage of 2.0 Volts and current of 20 mA to fire. From these numbers, one might estimate its [variable] forward internal resistance to be R = 2.0 V /.020 A = 100 Ohms.

What happens when one of the bulbs is replaced by the LED [Symbol:  ] ?  There were some dire predictions that the LED would burn out, because it would block the flow of AC current in the "back" direction, and thus get a maximum voltage of about 170 Volts across it every 1/60 second.  A counter argument would be that "current" rather than "voltage" is needed to burn up the LED, and there is no current flowing during reverse bias.  What really happens?

The LED did work well in the circuit, and did not burn out.  The current passing through the LED should have been over 100 mA, since the internal resistance of the LED is only slightly greater than that of one of the bulbs. Apparently, these LED's are tough little critters. Very interesting, Bill.

Walter McDonald (CPS Substitute Teacher; VA X-Ray Technician)
showed how to make an indirect measurement of the height H of a building, using trigonometry. He moved a distance D from the building, and measured the angle in the right triangle between the top and bottom, as shown:

| *
H | *
| *
| *
| *
| q *
It follows from trigonometry that
H = D tan q.
He illustrated the method by measuring the height of the ceiling, by measuring out D = 16 feet, and then measuring the angle q: [lying flat on the floor to get an accurate reading!] to be q = 30o. Thus,
H = D tan q = 16 feet * tan 30o = 9.2 feet.
Very good, Walter.

Notes taken by Porter Johnson.