Faceman's silly math trivias

[Faceman]

I'm impressed!

1. Hydrogen is the right answer indeed and of course was used in the original Zeppelins.
2. to give you a bit of the information I withholded: New York is about 50000 sqare kilometers. Even my tiny hometown Vienna is about 500 sqkm.
3. Yes I'd go with 1l/m^2 = 1mm rainfall for noticeable or with anything in the dimension of 10^0 for the sake of easier calculation.
4. We are of course disregarding that the explosion would be huge and have some effects that would prohibit our rainfall from occuring plain and right away.
5. Okay here comes my hastily put together calc.

So assuming I wanted 1l of rainfall in NYC

50 000 sqkm = 5*10^10sqm => we need 5*10^10l = 5*10^10kg water
1 mol H2O has 18g therefore we need
5*10^10/18*10^-3 mol = 2.7period*10^12 mol water
thus 5.5period*10^12 mol hydrogen
now we know that under normal pressure a mol of any gas has about 23l
that would make 1.27per*10^14l = 1.27per*10^2ckm
1.27per*10^2 = 4/3pi*r^3
3.05*10^1 = r^3
3.125 = r

We'd need a balloon with a diameter of 6.25km for NYC.
If you make the same calc for 100sqkm the solution is about 787m diameter so right there with your half mile.

[Faceman]

and in addition:

The LZ-129 Hindenburg was the largest aircraft ever to fly with a volume of about 200 000 m^3 actually filled with hydrogen.

Therefore it would have sufficed to create upon explosion (which actually happened in Lakehurst, NJ on May 6,1937)
78261l water.

2*10^8/23l= 8.695*10^6 mol hydrogen => good for 4.347*10^6 mol H2O
4.347*10^6*18*10^-3= 7.826*10^4kg or 78261l

Just enough for 1l rainfall over an area of 8ha.

[Faceman]

The next one is even easier because if I wanted to keep up with you guys I'd actually have to start thinking

A train leaves Chicago for Milwaukee and travels at a speed of 90mph.
There are 100 passengers aboard and the brakes have failed so there is no stopping the train.
In Milwaukee a platoon of angry math students with bazookas lying in wait to blow that horrid thing up.
If you had an UH60 Blackhawk, what is the maximum distance you starting/evacuation point should be from the track to make you able to evacuate all the passengers before the train reaches Milwaukee?
Your knowledge of the situation starts immediately after the train leaves Chicago (actually it is just running through because of the faulty brakes).


And a funny one (this one I stole from a book and did not think up myself):
A hunter and his dog are returning home. When they are in sight of their cabin (about a mile away) the dog starts running towards the cabin at 18mph while the hunter continues to walk at 3mph. When the dog arrives at the cabin he turns around and runs back to the hunter (who is now nearer tothe cabin) only to turn around again once he reached his owner and make a run for the cabin. He continues runninng back and forth between hunter and cabin until the hunter arrives at the cabin. How much distance did the poor dog cover?

And a free question: Give a proper justification of why Superman CANNOT lift a Jumbo (i.e. Boeing 747-400). Remember that Superman is ultra strong and invulnerable, so no: "Cos it would be too heavy even for him"

[Azred]

Originally posted by Faceman: And a free question: Give a proper justification of why Superman CANNOT lift a Jumbo (i.e. Boeing 747-400). Remember that Superman is ultra strong and invulnerable, so no: "Cos it would be too heavy even for him"

Because as soon as he begins to lift the plane the superstructure would not withstand the weight of the plane distributed over the small area of Superman's hands. The only effect would be him pushing his hands through the hull. [img]graemlins/saywhat.gif[/img] Worst case scenario--he lifts the plane, which subsequently breaks into two pieces. This is why Superman really isn't strong, but employs a modified form of telekinesis which effects only objects he can touch. [img]graemlins/beigesmilewinkgrin.gif[/img]

[ 08-20-2003, 03:28 PM: Message edited by: Azred ]

[daan]

And a funny one (this one I stole from a book and did not think up myself): A hunter and his dog are returning home. When they are in sight of their cabin (about a mile away) the dog starts running towards the cabin at 18mph while the hunter continues to walk at 3mph. When the dog arrives at the cabin he turns around and runs back to the hunter (who is now nearer tothe cabin) only to turn around again once he reached his owner and make a run for the cabin. He continues runninng back and forth between hunter and cabin until the hunter arrives at the cabin. How much distance did the poor dog cover?

Thinking up a mathematical formula would be more fun ( and not that hard ), but the answer is offcourse 1 mile [img]tongue.gif[/img]
He ran quite some distance, but s = distance from 0-point (startingpoint ) is still 1 mile, no matter how much times he ran back and forth [img]smile.gif[/img]

[Azred]

Originally posted by Faceman: And a funny one (this one I stole from a book and did not think up myself): A hunter and his dog are returning home. When they are in sight of their cabin (about a mile away) the dog starts running towards the cabin at 18mph while the hunter continues to walk at 3mph. When the dog arrives at the cabin he turns around and runs back to the hunter (who is now nearer tothe cabin) only to turn around again once he reached his owner and make a run for the cabin. He continues runninng back and forth between hunter and cabin until the hunter arrives at the cabin. How much distance did the poor dog cover?

The hunter moves at 1/6 the rate of the dog, so everytime the dog runs towards the cabin the hunter covers 1/6 of that distance. When the dog runs back, the distance from the cabin to the hunter must be split into 7 equal parts; the dog covers 6 and the hunter covers 1. So when running back, the hunter goes 1/7 of the distance from his previous position to the cabin.
What you get is a 2-step iterative process beginning with position 0 at time 0. At time 1 the dog is at position 1 and the hunter is at 1/6. At time 2, they meet at 1-((1/7)*(1-1/6)) = 2/7. Now the loop begins, because the dog runs 1-(2/7) and the hunter moves (1/6)*(1-(2/7)), placing him at (2/7)+(1/6)(5/7)=(17/42). The repeated second step shows the dog runs (6/7)*(1-(17/42)) and meets the hunter at 1-((6/7)*(1-(17/42)))=(24/49).
In the end, by continuing the process and adding the distances run by the dog: 1 + 2(5/7) + 2(25/49) + 2(218/343) + ... = 6 miles.

I presume daan is thinking of displacement, which compares the distance between the dog's starting and ending positions. [img]graemlins/petard.gif[/img]

[Faceman]

Nice work on the Superman question and thanks for the Telekinesis sidestep. If I look at it that way I think Superman can be my hero again. [img]smile.gif[/img]

calc:

Let's assume our hero's hands have an area of 300cm^2 each3
(this would equal circles about 20cm in diameter)
let's give him a big hand
A Jumbo has a mass of 300tons
therefore has a weight 3 000 000 N = 3MN
so
A = 0.06 m^2
F = 3*10^6 N
p = F/A
p = 5*10^7 pascal = 500bar (1 bar is common airpressure, 500 bar would be the waterpressure in a dephth of 5000m)

while there are material which can withstand that kind of pressure the Jumbo is probably not made of them (as air pressure drops with height there's not much need for that) and therefore Superman would probably just tear two big holes into the hull

IF he wasn't using his telekinetic powers


As for the hunter and the dog: Gotchya! [img]graemlins/hehe.gif[/img]

You almost had it Azred but then you (like everybody who knows his math and is not afraid to use it) stepped right into the trap.
The answer is easy:

The hunter moves at 1/6 the rate of the dog, so while he covers 1 mile the dog covers 6 miles. That's it

EDIT: Added Superman calc

[ 08-21-2003, 12:43 PM: Message edited by: Faceman ]

[Azred]

At least I arrived at the correct answer, even if I did take the long route. [img]graemlins/petard.gif[/img]

Other Superman stumpers:
If he were to take off flying at top speed from a stationary position on the ground, the vacuum created would not only knock everyone in the vicinity off their feet, the subsequent sonic boom would probably deafen them and knock them backwards, as well as blow out glass panes. [img]graemlins/wow.gif[/img] Superheroes and physics...a risky combination.

[Deejax]

Originally posted by Faceman: A train leaves Chicago for Milwaukee and travels at a speed of 90mph. There are 100 passengers aboard and the brakes have failed so there is no stopping the train. In Milwaukee a platoon of angry math students with bazookas lying in wait to blow that horrid thing up. If you had an UH60 Blackhawk, what is the maximum distance you starting/evacuation point should be from the track to make you able to evacuate all the passengers before the train reaches Milwaukee? Your knowledge of the situation starts immediately after the train leaves Chicago (actually it is just running through because of the faulty brakes).

Like, umm, i think I'm missing some information:
How fast is a blackhawk;
how many people fit in a blackhawk;
How long does it take to load/unload those people from the blackhawk;
What's the distance between chigago and milwaukee;
What is the range of a bazooka;
How many tries do angry math students need to hit a moving train with a bazooka;

You do make these questions difficult for us foreigners [img]smile.gif[/img]
I'll do some searching and will get back to you.

[Faceman]

I withhold vital information to make this easy things a little bit more difficult
And I actually use US related stuff to ease it up. I'm not from the US myself and have to look for the info as well as you.

But for the sake of an easier calculation please assume that Chicago-Milwaukee is a straight (which is not exactly true) and for the math students: They actually can't really figure out how to launch that bazooka but they have cunnigly chosen to set camp right at the tracks and probably will explode along with their bazookas when the train hits them.

[ 08-21-2003, 12:48 PM: Message edited by: Faceman ]