First, a disclaimer. This puzzle came from the Car Talk program on NPR, so listeners of that program are not allowed to give the answer!

You have two rather tall telephone poles, each 100 feet tall.

You have a 150 foot long rope strung between the tops of the two poles, and the poles are separated by a distance that allows the lowest point of the rope to be 25 feet above the ground.

How far apart are the two poles?

(Hint: the curve the rope makes is called a catenary.)

Be sure to show your work when you solve this...


tanstaafl.

[rot13]
Gurl'q unir gb or gbhpuvat. Abg zhpu bs n pngranel, ernyyl.

Fbeel, Qbht. Gbb rnfl. V qvq nyzbfg guvax nobhg ybbxvat sbe n sbezhyn sbe pngranevrf, gubhtu. Naq V'z fher vg'f zhpu yrff boivbhf jura erynlrq benyyl.
[/rot13]

Oh, and since it always takes me too long to remember the exact correct syntax:

tr '[a-zA-Z]' '[n-za-mN-ZA-M]'

Are we supposed to account for stretch in the rope due to gravitational affects?

Sorry, Doug. Too easy

Damn, Bitt! I should have waited until you were off the bbs before posting the puzzle!

tanstaafl.

Too bad I can't make the foreground and background colors the same so that it can be hidden. Maybe I'll ROT13 it for everyone.

I guess that answers my question.

But speaking of catenaries, I remember the first time I ever heard that word. My dad went to a trade school type thing for land surveying around the time the Batman series (with West) came out on TV. The whole class begged the prof to be let out so they could all go watch the premiere. The prof let them go, since I think he wanted to see it, too. That episode showed Batman and Robin climbing up the side of the building, and there was a catenary in the rope that was supposed to be dangling straight down. Apparantly no-one ever asked the prof to get out of class to watch it after that.

rot13

Man, a Solaris guru like yourself using such a weak form of "encryption." Couldn't you at least use quadruple rot13 next time?

Well, excuuuuuuuuse me!

-----BEGIN PGP MESSAGE-----

Version: GnuPG v1.0.4 (OpenBSD)

Comment: For info see http://www.gnupg.org



jA0EAwMCGNFuBVPKgJxgycABjMfaNLfXP4Xykl0ODY5IsMivKwxZQQFRlxAuyYKR

XLpuS8WDKqd1IuHhB3yxlx0DYsX4av+nyyKbeyHZqYmhddxZe/+DOK7x2W2+JQNY

uglhKUs70gXyKwh3BFeZ5xiHEKN7eyjOb7p4yQkXGNdCvgJSU+44ZH72qeLjQ0es

A4Ew7Gk01BzNb2pGjtHBe7Kxlcao6LS93tcjUYDjBDAcoZr0ecz9w57EU3d+I0c7

bEVlfd6R6bgdQN/Xabt/Lm/QHQ==

=O0t1

-----END PGP MESSAGE-----

Symmetric cypher. Passphrase is:

Tony's a big know-it-all

Sheesh, tough crowd!

Maybe I'll ROT13 it for everyone.


Nice. That'll keep up the suspense for a little bit. I expect muzza will have a go at it...

SInce that one was too easy for you, try this one. It is another Car Talk puzzler.

A truck driver (tractor-trailer rig) has those big fuel tanks under his running boards. Each tank is two feet in diameter, a bit over four feet long, just about exactly 100 gallons per tank.

His fuel gauge has been broken since 1997, so he sticks the tanks with a yardstick to get an idea of how much fuel he has on board. He knows when the yardstick shows 12 inches of fuel, the tank is exactly half full, or 50 gallons.

But how much fuel is in the tank when the yardstick shows six inches of fuel? Obviously it is going to be considerably less than 25 gallons...

Now, this trucker doesn't remember how to find the area under a curve using calculus, so he came up with an elegant way, using simple math and household items, of getting a close approximation of fuel on board when the yardstick shows six inches of fuel .

How did he do it?

tanstaafl.

Is there going to be a lot of wasted Morton's Salt or Quaker Oats involved in this?

Is there going to be a lot of wasted Morton's Salt or Quaker Oats involved in this?

Not a bit!

Hmmmm..... bit. There's a pun lurking in there somewhere, I'm just too dense to see it.

tanstaafl.

They are obviously right next to eachother since the rope hangs down 75 feet and is only 150 feet long.

Simple way to measure the amount of fuel: when it gets down to six inches, go to the gas station and fill it up. Note the amount of gas that it takes and subtract that from 100 gallons.

What's next?

What he will need is some scales, and a circular piece of something. If he has very acurate scales, this could be a piece of paper. Weigh the whole piece and let that weight represent 100gallons. (Say this weighs 300 carat (you introduced the use of irrational measurements))

Then cut off a "cap" of the circle by cutting a straight cut which is a 1/2 radius from the edge. Then weigh this piece, and that will tell you the remaining amount. (Say this weighs 15 carat, you'll then know this is 1/20 of the tank or 5 gallons)

<a href=http://empeg.comms.net/files/165974-gallons.gif> Picture </a>

The only thiung I learned in chemistry class is that chemistry people don't like mathematics. So, when we wanted to find the integral of a function, we drew the graph on a piece of paper, and cut out and weighed the graph to find its area.



Marius (Escort Cab + Mark II)

picture

~15 gallons

or ~20 gal. if the cylinder has domed ends

or ~9 gal if it is a 48 inch sphere with 6 inches left

Does it involve constructing a lure to trap someone who is good at calculus?

Put a 5 gallon jerry can in the back of the truck, drive round 'till it's empty and put the five gallons in, then drive to the nearest gas station and fill it up to the requisite depth: The answer is what shows on the pump +5!

I wouldn't exactly call lab scales accurate enough to measure a small piece of paper "household".

[rot13]
Ur hfrq n fvk vapu qvnzrgre pbssrr pna, qerj n qvnzrgre npebff gur (genafyhprag) yvq, znexrq bss bar cbvag svir vapurf va, naq gura qerj n crecraqvphyne yvar. Sebz gura ba, rirel gvzr ur znqr pbssrr, ur'q gvc gur pna ba vg'f fvqr naq ebyy vg fb gung gur yvar jnf cnenyyry gb gur tebhaq. Jura gur erznvavat pbssrr jnf ng gur yvar, ur jrvturq vg naq qvq gur zngu. (erznvavat/fgnegvat * 100 tnyybaf). Ur qvqa'g rira arrq gb erzrzore ubj zhpu pbssrr ur unq gb fgneg jvgu, fvapr vg jnf nyernql cevagrq ba gur pna.

(Onfvpnyyl wnar'f zrgubq hfvat ubhfrubyq vgrzf)
[/rot13]

I guess "welding equipment" is houshold items in the US? In that case, he could use a sheet of metal... or even a wooden disk... or crack a kitchen plate...? Or the platter of an old Harddisk? (would perhaps be a problem with the hole in the center?)

Marius

I just love how rot-13 looks like Klingon.

Have you been applying for that University Klingon Prof posting?

-Zeke

Ahh, a Klingon translator: http://www.pflock.de/rot13.htm . This has helped make sense of several posts for me.

No, I'm crap at calculus.
http://grapevine.abe.msstate.edu/~fto/tools/vol/index.html

Now, this trucker doesn't remember how to find the area under a curve using calculus, so he came up with an elegant way, using simple math and household items, of getting a close approximation of fuel on board when the yardstick shows six inches of fuel .

How did he do it?

He used his household computer to post the question to a geeky BBS?

He used his household computer to post the question to a geeky BBS?

None of the above.

All he needs to do is take a glass jar (like a mayonnaise jar) and using the same yardstick he used to stick his fuel tank, measure and mark the round end of the jar at 1/4 the diameter. Now, put water in the jar and by trial and error add enough so that when the jar is lying on its side the water level is at the mark.

Next, and here's the clever bit, stand the jar back on its end in the normal upright position. Measure the distance from the bottom of the jar to the water level, and figure this as a percentage of the total height of the jar. For example, the jar might be nine inches high, and the water level at 1.4 inches. 1.4 divided by 9 = about 15% filled. (I am not saying that these would be the actual numbers, but are just given as an example).

The same ratio would hold for the fuel tank. If the tank held 100 gallons, then when the fuel level was one quarter of the way up the diameter of the tank, there would be about 15 gallons in it.

I have to confess that when this puzzle appeared on Car Talk, I was unable to solve it. But I certainly appreciated the ingenuity and intuitiveness of the people who did.

tanstaafl.

Every tricorder (a.k.a. *ix machine) contains Klingon translator. It's rather simple, actually, and Bitt provided it in his first posting in this thread:

tr '[a-zA-Z]' '[n-za-mN-ZA-M]'

No, I'm crap at calculus.

I'm still a bit confused by responses like this. Even if you do it the hard way, it only needs simple trigonometry, not calculus (in UK terms, it's a GCSE question not an A-level question).

Consider a circular cross-section of the tank. The fuel comes halfway from the bottom to the centre. So by looking at the triangle formed by the centre of the tank, the point on the fuel's surface directly below the centre, and the "shore" of the fuel where it touches the edge, you can work out the angle at the centre, i.e. the angle that one side of the fuel subtends at the centre. It's the arc-cosine of 6/12, or 60 degrees. So the whole surface of the fuel subtends twice that, or 120 degrees, at the centre of the tank.

So what area of the cross-section is taken up by fuel? Well, it's the area of the 120-degree pie-slice shape, minus the two triangles. In other words, it's 1/3 of pi*12*12, minus twice 6*(12 sin 60). If you divide that by the total area of the cross-section, pi*12*12, you get the fraction of the total volume occupied by fuel. Multiply that by the capacity and you get the volume of remaining fuel.

Now all this needs a calculator, of course. But if the trucker's smart, he'll do this only once and calibrate his dipstick in actual fuel units.

Peter

Just reading this thread is making my head hurt. I think I'd rather get someone to fix the fuel gauge and be done with it.

You all are forgetting that the yard stick takes up volume, thus raising the level of the gasoline and giving a false reading. The solution I gave is the only easy and accurate way. The jar thing I'm not really going for. What are the chances of finding a jar that is proportionately equal to the tank?

The proportions don't make any difference. The only thing that you're calculating is the area described by the arc of a circle and the segment connecting its ends as a fraction of the total area of the circle. The depth of the cylinder is irrelevant in determining this.

Of course, you have to contend with the non-cylindrincal dissimilarities of both the tank and the jar. But we're not looking for precise measurement. Your method wouldn't provide that, either, as it's difficult to make the level of a tank the same after each filling. It also doesn't provide for arbitrary determination, which is certainly useful, if not actually required in this problem.

Most gas tanks have the ends of the cylinders rounded out while most jars are rounded in. Combine that with the yard stick giving a false reading and you have an inaccurate solution.

If the tank is exactly 100 gallons then accuracy shouldn't be a problem using my method. Plus, this method will work with any shape tank, making it more applicable to the real world.

Yeah, and it'd be real convenient when you get to an amount you haven't measured before.

BTW, since you're being annoyingly pedantic, six inches of a yardstick that measures a quarter of an inch by an inch would displace about 0.0065 gallons, which is far less than the precision you'd be able to get out of a pump. In addition, you're not taking into account the fuel you'd burn driving to a gas station. Or maybe you meant that you should call a truck to tow you to a gas station.

In all honesty, the levelness of the tank is likely to cause as much error as non-cylindrical abnormalities.

And you expect him to drive to the store and buy a jar?!

What do you think he pees in?

What do you think he pees in?

A High-flow toilet, smuggled in from Canada, of course.

-jk

What do you think he pees in?

depends...

What do you think he pees in?

depends...

No, that would be her