Steven S. Holland

A few months ago I had seen a thing for a clock with a smith chart as it’s clock face which I searched and can’t seem to find a link for that. Regardless, I have a cheap black wall clock that I picked up while cleaning up my room and thought that the clock face looked to be about the right size for a printed out smith chat, and I was right:
So I took a screw driver and pushed in two tabs on the back of the clock that hold in the clear plastic over the face of the clock, and removed the clear face:
Next I wanted to get the paper clock face out to use as a cutting guide on the smith chart, so I cut from the edge to the center hole so I could remove the face without trying to remove (and likely break) the hands of the clock:

Next, I took out the paper face, and aligned that on the smith chart so the smith chart would be more or less centered (with a perfect match at the axle of the clock arms!). I put tape on the edge of the clock face so it wouldn’t shift from the position I centered it on:
Then I cut out the circle on the smith chart around the edge of the clock face — the cutout needs to be the same size or smaller than the clock face or it won’t fit in the clock:

Then I cut out the circle in the center of the chart (shown started above) and cut a line from the outside of the smith chart to the center, again, so I can get it on past the arms of the clock. In retrospect, I could have cut a small line out from the center as long as the back end of the seconds hand, and then put the long end of the arms through the hold and then slide this slit down over the longer back of the seconds hand, thus avoiding cutting a line from edge to center, so if you are doing this, i recommend not cutting all the way from the edge.

I then put the smith chart in on top of the old clock face, and lined up the real impedance line (the horizontal line on the chart) with the 3 and the 9, so that a perfect short is 9 o’clock and a perfect open is 3 o’clock. I taped it down on the back side in a number of places to hold it snug to the clock face, to avoid it shifting around. I debated putting an extra circle in between the clock face and the smith chart to make sure the numbers don’t bleed through — they don’t unless under strong light, so I don’t think it’s necessary.

Here is the finished product!
So now, 12 is at Z = j, 6 is at Z = -j, and so forth. I am debating putting the clock numbers on this, but so far I am thinking I like it the way it is. I can now tell time in either standard, military, or impedance.

Some facts:
-the seconds hand traces out a circle at the tip that has a VSWR of about 12, the hour hand a VSWR of 2.8, and the minute hand VSWR = 9 or so.
- Between 3 and 9 o’clock each day, the hour is capacitive, while between 9 and 3 o’clock the hour is inductive. The minute hand, obviously, spends half of the hour between 15 after and 45 as capacitive, and the rest of the time the minutes are inductive.
- At any given hour (and impedance), moving 6 hours around the clock gives the admittance of the impedance. Same with the minute hand, only moving 30 minutes around the clock instead.
-9 o’clock is a voltage minimum, whereas 3 o’clock is a voltage maximum..

,,,I could go on, but I think I beat that one to death. Make your own smith chart clock!

De Moivre, famous mathematician, is known to have predicted his own death. After finding that he slept 15 extra minutes with each night, he solved a series called an arithmetic progression, which is basically a series where the difference of each successive term is constant. example:

1 + 3 + 5 + 7… is an arithmetic progression because the difference between all terms is two.

Anyway, by summing this series up he figured out the day he would sleep for 24 hours, and declared that as the day he would die. He died on that very day – November 27th, 1754.

My best solution is 2 weighings. I figure as this:you take 6 of the balls, and weigh 3 on each side. if they weith the same, then weigh the two remaining balls to find the heavier ball.

If they are different. then weigh two balls from the heavier set of three balls. If one is heavier than the other, there’s your heavier ball. if they weigh the same, than the 3rd, unweighed ball is your heaviest ball.

If you have 8 billiards balls, and one of them is heavier than the other 7 – given a balancing scale (where you compare the weight of something on one side of the scale to the other) what is the fewest number of times you need to weigh in order to find the heaviest ball?

I thought about this and came up with what I think is the right answer. I will post the answer tomorrow morning, Wednesday.

Ok, so my last post talked about the hits I have been getting from the xckd comic that is similar to a challenge problem I posted. So, I solved the same problem as on the xkcd comic:
link!

This solution is heavily based on the work I did for this solution, so you should review this before looking at this one if you haven’t seen this solution before, as it explains what the impedances shown below are and how I reasoned them out.

So here is my solution. I took a block of resistors around the two points, and the following rewriting of the circuit follows:
This is a true representation of the circuit. The T-Delta circuit transformation is a standard circuit theory technique. From here the circuit then looks as shown below, with the circuit theory solution then of the thevanin equivalent resistance.
I haven’t computed the R thevenin yet, which I will at a later date. However, knowing from this solution that Zin = (2/3)R, the new equivalent resistance can be solved by substituting this into the T-Delta equations and then into the final equation. Again, this will be posted when I get around to doing this.

Please let me know if you find any errors in my work…

Solution #1 (my solution I posted on the chalkboard)

First I stood there looking at the picture, and thought that if it is infinite in extent, all the square grid is essentially a repeating pattern of squares, then the input impedance looking into a side of one of the squares must also be periodic and symmetric in all directions. Therefore I said that if inside one of the squares, looking toward any of the sides of the square must be the same impedance, which I called Zin. As a result, the input impedace looking out of the box on any side is the same as looking into any one of the sides. Finally, I said if I look into the box from any one side, I can say the other 3 sides look like Zin in series all in parallel with a resistor “R”, as shown below:

So from the last sketch on the right, I believe this first step to be the trick — because from here on out is simple circuit analysis theory. Another way to think of it is that since it’s periodic, addind or subtracting a box should not affect things, so remove the box to the left of the box and then look along the Zin path.

So next, redraw the circuit like a normal circuit and solve for Zin as shown below:
This is not the answer, just yet. This is what the resistance looks like looking out of or into one side of the box.

In order to get the actual thevenin resistnace, now solve for the inpedance looking into the terminal set a-b, as shown below:
SO, the answer turns out to be that Rth is equal to one half of R, as shown above. This was the mathematical solution.

Solution #2 – Here’s the cool, clever solution! (that I did not think of but that I found on google — what can’t you find on google?!)

This solution is so much simpler. It requires not thought about periodic impedances or input impedances looking this way and that way. It relies on symmetry, Ohm’s law and Kirchoff’s Current Law.

The first step here is to look at node “a” and “b” separately:
The next step is to apply a current source of 1A to terminal “a”, leaving terminal “b” alone. The current can be setup flowing into node “a”, as shown. The consequence is that since this huge grid is symmetric and infinite in extent, the current branches off into 4 equal currents, or 1/4 of an amp. Also, the other side of the current source can be said to be placed at infinity.
So, from the source at “a”, 1/4A will flow into the branch between node “a” and “b”. Now take away that source and apply another 1A source to node “b”, only this time flowing out of the node:
Once again 1/4A flows through each branch. Using the superposition theorem, the total current when both sources are applied will be the sum of the current from the source at “a” and the source at “b”, giving 1/2A flowing between node “a” and “b”.
Since one end of each current at infinity is moving 1A through the node at infinity (at node “a” source 1A is extracted whereas at node “b” source 1A is injected) then the two ends of the current sources can be thought of as connected, and therefore replaced by a single 1A current source.
Now, by definition of the thevenin equivalent circuit theorem, driving two terminals with 1A of current and then dividing the voltage that developes across the terminals by 1A, you get the equivalent resistance. This is shown above. The voltage is equal to the resistance “R” times the total current flowing through it, and diving this voltage by 1A gives the final equivalent resistance.

What an elegant solution. It takes explaining, but math-wise it is simple and it is very intuitive, using addition, basically.

So, I was walking through the engineering building here and came accross a chalk board that often has random stuff on it. Its basically a chalkboard for whatever you want to write. Sometime’s there are quotes written there, sometimes tick-tac-toe games, and sometimes some funny drawings. Today there was a brainteaser — specifically a circuit analysis brain teaser. Here’s a sketch of what was on the board:
This is an infinite grid of resistors (all of value “R”) – the elipsis implying that in all directions the circuit continues indefinitely, and the n is just saying this goes on til some “nth” resistor at infinity.
TASK:
Find the thevenin equivalent circuit across terminals a and b (essentially whats the thevenin resistance looking into those terminals).

I would like to be able to say this is an incredibly hard problem that I found the solution to, all Good Will Hunting style, but its just a brainteaser for circuit theory — there is a trick to it, and its neat, but other than that it’s not really anything all that impressive.

It took me a few minutes of thinking but I came up with a solution and it is now posted on the board.

I will state up from that there is a very easy, intuitive, almost no math solution to this problem. There is also a second, more complicated, mathematical solution. I was not clever enough to think of the elegant simple solution, and instead found the more complicated solution. Wasted work, if you ask me. I only found out about the simple solution because when I got back to lab I googled the problem to see if there were any solutions i could check my answer off of (and to see if there were more solutions). Thursday I will post both solutions. Best of luck.

EDIT: I drew only one “cell” of the grid, and drew elipsis to show it goes on forever in all directions, but to make it clear, the overall topology is a huge square grid that is infinite in extent in all directions, as shown below: