1. How can a train ascend a 1-mile radius hemisphere, without exceeding a 4% grade? The train track must spiral up the hill as far as it can, at a 4% gradient. About 25 miles or so, up to point "X". And how far is that, exactly? Where is this "point X"?

Let's look at this from the standpoint of starting down the hill. The track will go STRAIGHT DOWN the hill as fast as it can - but it starts out on the level, and gradually starts getting steeper, as it descends further along the Great Circle. How far down can it go? It can go down to a point where the gradient reaches 4.0%, and then it has to curve over to one side, and start spiralling. That defines Point X, in terms of its (r) and (Z) coordinates. (It does not tell us about the angle, theta.)

How can we describe this point X? It's a point where the slope (the tangent) is at 4 degrees. There, the radius (from the center of the hemisphere) is still exactly 1 mile, and the vertical elevation above the plain is exactly - 1 mile x cosine of (arctan 4%). About 0.999200958 mile. And, this is about 4.21894 feet closer to the plain, than the top of the hill. In other words, you only descend 4.21894 feet of elevation, before it has hit the 4% gradient. That's the view from the top. And this point is barely 211.2 feet from the top, before it has to curve.

Now, starting from the bottom, every mile you ascend and travel, at a 4% grade, the horizontal distance is - cos (arctan4%). So when you travel one horizontal mile, the vertical ascent is 0.040000 mile, and the amount of track needed is - 1/cos (arctan 4%) - or about 1.00079968 miles

When you need to spiral up to ascend a vertical distance that is exactly 1 mile x cos ( arctan 4%) - the amount of track needed is exactly 25.0000000 miles, to get UP to Point X. The value of cos (arctan 4%) cancels out, in both parts of the math - - - so you never have to know the exact value of this quantity.

What is the distance from point X to the top of the mountain? Well, it is quite close to (4% of 1 mile), or 0.04 x 5280, or 211.2 feet. That is 4% of a mile. But the angle, in radians, is NOT 0.0400, but 0.039978687. That is the arctan of 0.0400. If you go in a straight line that would be (211.0734 feet) But since we are travelling along an arc with a true radius of 5280' - - the actual mileage from point X to the top, along the arc, is very close to 211.0875 feet.

The TOTAL is quite close to 1 mile x [(1/.04) + (.04)], or 25.04000 miles, or 132,211.20 feet. That is a very good estimate. Actually you don't need 132,211.20'; 132,211.087 feet will be enough, so 25.0400 miles would be over by about 1.35 inches. But since this isn't even 1 ppm over, it won't cost Egbert too much.

Nobody buys rail with an accuracy better than 1 PPM. To do that, you would need a very tightly controlled temperature-controlled rail-yard.

Problem 1.b. If you start on the exact west side of the mountain, heading north, from what angle do you approach the summit? In other words, at what angle is "point X"? I have no idea. My best guess is: About (XX) degrees. But that is a very crude guess. I'll allow you guys who can program computers to compute trivia like this, tell me what the answer is. I'll post the best answer that 2 people can agree on.

Problem 2. When Egbert has travelled 2.000 miles, he is already very close to the goal. In fact, this hypotenuse is 2.00249844 miles, so he is only 0.00249844 miles from the goal, or 13.19176 feet. The other fellow has 528 feet to go. Egbert continues at 2 mph, and will arrive in 4.4972 seconds. His brother must go 528 feet in this time, or 528 feet/4.4972 sec, or 117.41 ft/sec, or 80.05 mph. It is OK to round this off to 80 mph. Or 80.1 mph... so he won't be late.

When you are doing this in your head, it is OK to simplify a bit, that the hypotenuse is 2 miles + 1/400 mile. Thus the ratio of the distance Egbert has to go, versus the distance his brother has to go, is 1/400 versus 1/10, or 1:40, so the solution must be 40 x 2 mph. This answer is within 1%...

Problem 3: For the problem with 4 doors, at each of the 4 switch locations, install a DPDT switch, and connect as show in this diagram:

Pos. 1:        Pos. 2:

(a)O  O(c)     (a)O - O(c)
    \/        
    /\
(b)O  O(d)     (b)O - O(d)

The contact bar would connect diagonal contacts or adjacent contacts.

In the following diagram, the load is energized; throw any switch and de-energize; throw any switch again and load is re-energized

               0-------0  0-------0 - 0----------0
Source -------/         \/                         ----load
                        /\                        /
               0-------0  0-------0 - 0----------0

The physics prof didn't know how it worked either; I had worked for an electrician during the summer and found out about it.

Richard Hensel,
Non Impediti Ratione Cogitationis
Lexington, Ohio
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*** Hello, Richard,
Well, that is a great solution. I did not know about "crossover" switches. Of course, one can take a DPDT switch and wire it to act as a "crossover", but many hardware stores do not sell DPDT switches, to mount on your wall. YOUR elegant solution uses about 1/2 as much wire as mine does! And YOUR solution can obviously be extended to n = 5 or more. Thanks for writing! / rap

Note, it's true that a room that's 100 feet square, would need about 1000 feet of heavy-gauge wire. And the I x R drop of several amperes through 300 feet of wire is NOT small or trivial. Using light-gauge wire to drive a relay would begin to make sense.

And, if the number of switches goes past 4, using SPST switches which drive light wires, which drive exclusive OR gates, which drive a relay, would become effective. Okay?

Problem 4: One good way to find the square root of 156 is to recognize that 12.5 (squared) is 156.25. Now, 156 compared to 156.25 is 0.16% low. So the square root of 156 must be slightly less than 12.5. About 0.08% less than 12.5. That will be very close to 12.490.

One good way to get VERY close to the square root of (N) is to divide (N) by a number which is close to (sqrt N) - let's call it (n). You then get a result: N / (n) = n'. Then the square root of N is very close to (n + n')/2. This is because (A + b) x (A - b) = A (squared) - b(squared). If b/A = .001, then (n + n')/2 = (sqrt N) with a precision of 1/2 ppm. If b/A = .000001, then (n + n')/2 is equal to the square root of N with an accuracy better than 1 part per TRILLION.

In this case if we divide 12.4900 into 156, the quotient is about 12.48999200. This is easy to do in your head, as it's just a bunch of chained additions, multiplications, and subtractions at the 4-digit level. So the square root of 156 must be VERY close to ( n + n')/2 = 12.48999600. Well within 1 ppm. Actually, about 0.003 ppm.

If you then want to divide THIS into 156, you can pin down that the solution is closer to 12.48999599680.

Some calculators will tell you that the (sqrt) 156.0000 is 12.489995995, but this is an example of an incorrect round-off. It's 12.489995997 (if you round off to just 11 places) or 12.48999599680 or 12.489995996797.

THIS solution is down below 1 part per TRILLION. Even for space travel, we don't need accuracy better than that.

So this is just to remind you that if you need a square root, you don't need a calculator that can do square roots. You can get accuracy well below 1 ppm with just a calculator that can just do DIVISION, and the right techniques. Or, you could use a stick to write in the sand. Or, you could do it in your head....

Problem 5. When you drop an object (such as a plumb bob with a point) about 16 feet, it will fall to the ground in about 1.00 second.

The radius of the earth is ABOUT 4000 miles, or ~ 21,000,000 feet, or ~ 250 million inches. The velocity of the ground, at the foot of the ladder, is about 2 pi x 250,000,000 inches per (86400 seconds) (i.e. - one day).

The velocity at the top of the ladder is 2 pi x 250,000,192 inches per (86400 seconds). The relative speed is that the object atop the ladder is moving faster, by 2 pi x 192 , inches per 86400 seconds. We do NOT have to know the actual radius of the earth - or the speed - we just need to know that the top of the ladder is moving faster by 2 pi x 192/86400, in terms of inches per second.

Now, you don't actually have to crunch those numbers, with a slide rule or calculator. Because there are about 7 factors of 2 in each of the numerator and the denominator. After you cancel out all the common factors, the final answer is about (pi) / 225, or 1/72, or 13.96 milli-inches. Say, 14. That's well within 1% or 0.1%, actually.

THUS: If you drop an object, which is moving EAST, FASTER than the ground, then in the time the object falls 1 second, the object moves FARTHER EAST, farther than the ground. It will hit the ground in a place that is offset about 14 milli-inches east of the point it touched, when it was LOWERED to the ground. (In that one second, both places will have moved about 17,900 inches... ) (This is a resolution of about 1 part per 10 billion, comparing the RELATIVE speed to the ABSOLUTE speed.)

If the height of the ladder is 16 x (M) feet, then the distance will be 14 mils x (M) exp 3/2. So if (M) = 4, and the ladder is 64 feet high, the offset distance would be a definitely noticeable distance of 112 milli-inches. Of course, you would have to avoid all winds, and you would have to have a release mechanism that imparts negligible speed to the falling object. And the step-ladder can't wiggle. OK, drop it down a well.....

This effect is not EXACTLY the same as a Coriolis force, because all forces are oriented directly toward the center of the earth. All MOTION is referred to the plane that is moving "EAST" at about 17,000 inches per second. If you do this much more than 1.0 seconds, you would find the rotation is annoying...

This just goes to show, that when you drop something, it does not just fall DOWN. It falls in a sort of elliptical orbit, around the center of the earth (in the absence of wind or friction or any other influence). So, NO, when you drop something, it does not just fall "down". It falls in some kind of orbit. And, if you measure closely, you can tell the difference. Thanks to "Goldie", Maurice Goldwater, from 40 year ago.

Best regards! Comments invited. / rap

I believe we got an answer to Problem 1b. Israel Schleicher, by e-mail, and Ermi Roos of Miami Florida sent in simple and elegant solutions that the train circles a little more than 6 times, and both agree that the train approaches the Summit from an angle of 32.7347 degrees north of west. If you argue that you have a solution that is different from 32.7 degrees north of west, you are probably wrong.

Details to follow. / rap