monk

[reverse]

I don't enjoy math sorry.

 

But I seriously do remember this problem in a slightly different variation from ages back.

 

Hey Callipygous,

What was the main hitch you saw in solving this one?

[Callipygous]

I don't enjoy math sorry.

 

But I seriously do remember this problem in a slightly different variation from ages back.

 

Hey Callipygous' date='

What was the main hitch you saw in solving this one?[/quote']

 

 

figuring out how many degrees of the cirlce its still in contact with.

 

 

hmm.... maybe if i find a tangent line and then use vectors to find the difference in the angle of the tangent lines.... /chinstroke

[Martin]

figuring out how many degrees of the cirlce its still in contact with.

 

 

hmm.... maybe if i find a tangent line and then use vectors to find the difference in the angle of the tangent lines.... /chinstroke

 

idea of using tangent lines sounds like a good one!

 

 

the actual math part doesnt have to be very advanced (I should think that, once one has the picture, highschool trig should be more than enough)

[reverse]

oh I see,

the problem I saw was slightly different.

it was string not elastic..

 

I see what you mean.

 

It's about triangles and circles.

 

that ratio of 64k to 0.5, is going to make for an interesting tangent.

 

I'm still suspicious that this may not be a simple math trick.

 

well if it was just math,

you can get the length of a side by adding the ladder on to the radius which is related to the diameter through the Pi ratio?

and you know the length of your other side already.

so Pythagoras told us how to get the last side length?

(and also how to find the angle at the hub or core of the planet...

which we double for the other side of the ladder)...then subtract twice the shortest sides of the triangle form the % dia...???

 

this is way too sluggish.

there must be a shortcut.

 

 

that's it...I dont enjoy math,I'm going to build something.

later.

[reverse]

ok found out more about this one.

it's by langman from 1935.

 

there is also a paradox named after him.

[Callipygous]

reverse, that actually sounds much easier than my way... nice.

[Dr. Casey]

6 meters, isn't it? I mean, if the stepladder is half a kilometer, and you stretch the rubber band 500 meters, you'd be stretching it far enough to eclipse the distance covered by the difference in length between the altitude and the diagonal measures of each vertex of the stepladder.

[reverse]

are you calling the surface of the planet a straight line?

 

can you do that?

[Martin]

are you calling the surface of the planet a straight line?

 

can you do that?

 

no you cant do that. it is important to the problem that the surface is curved (radius 6400 km)

 

I think Casey is right or within a meter or so of right.

 

How did you do it Casey?

[Martin]

so far, Casey is the only person who has offered a figure and he says 6 meters

 

if anyone else, Calli or reverse (?) would like to offer a solution please do, it might be closer than Casey's 6 meters.

 

please show your calculation (I dont follow what Casey said, or how reverse's plan would actually go in a real calculation either)

[Callipygous]

i get 3.3 meters, using reverses method. im gonna work through it again now. (think i have rounding problems)

 

now i got 8.3329 meters... i think thats more accurate than the first one, but how close it is i have no idea : P

[Martin]

good, so we have 6 meters, 3.3 meters, and 8.3 meters

 

no one has shown a calculation yet, though

 

I will show a calculation and you can critique it and refine it, if you want, so as to improve the accuracy---if there is some obvious way to do this.

 

 

I draw a picture of a circle with two tangents intersecting at a point (which is where you are at the top of the ladder)

 

I denote by Theta half the angle that the rubberband is off the ground

 

 

to be real detailed and explicit, I draw two lines from the center of the earth, one of length R from the center out to where a tangent touches, and one of length R+H from the center out to you at the top of the ladder,

and Theta is the angle between those two lines where they meet at the center of the earth.

 

[math]\cos \Theta = \frac {R}{R + H} [/math]

 

(the tangent forms a right angle with the radius so we can use trig for right triangles)

 

1. by pythagoras, the length of one of the tangent lines from where it touches the radius up to the top of the ladder is

[math]\sqrt { 6400.5^2 - 6400^2} = 80.00156 [/math]

 

2. the length of that segment of rubberband before it gets stretched is the angle Theta (in radians) multiplied by 6400 kilometers

so I just do arccos of 6400/6400.5 on the calculator (making sure it comes out in radians) and multiply that by 6400, and I get 79.99739

 

3. there are two segments (the picture is symmetric) and by subtracting those two numbers I find that each segment has been stretched 4.164 meters.

 

So it looks like Callipygous answer is right! Twice 4.16 is about 8.3, the total stretch is about 8.3 meters!

 

I hadnt actually worked it out, just knew the rough order of magnitude what it should be. Didnt realize til I worked it out that your 8.33 meters answer was right on.

[Martin]

how about you dig up/think of a problem for us?

 

 

 

BTW Casey is new (only registered a couple of days ago)

Welcome Casey

[ski_power]

Since I dont have an inverse calculator, I'll put down the method over here.

1. x = cos-1(64/64.005) [x is in radians]

2. c = (2 pi - 2 x)* 6400

3. d = 64000 * tan(x)

4. d = 2 * d

5. a = c + d.

6. cout<<a

7. a1 = 2*pi*6400

8. a2=a-a1

9. cout<< a1<<a2

 

I hope the above is self explanatory, if not, then here goes.

x is the angle bet. radius to tangent and distance from centre to highest point on ladder

2pi -2 x is remaining angle in radians

c is circumference of the portion excluding the 2 tangents.

d is the length of the tangent. 2d is for the 2 tangents.

a is the total final length of rubber band.

a1 is circumference of the standard spherical earth given in question.

a2 is the difference a.k.a stretch.

 

Now, just someone to calculate using these values...

[ski_power]

darn it took so much time to work and type it out, martin already put the solution...

[Callipygous]

i did your step one to find the length of the elastic off of the ground, then did sin with that length and the hypotenus for theta in the middle. doubled that and subtracted it from 360 so i could find how much of the circumference the string was still touching. that +2 times the length of the tangent, - the circ. of the the planet= stretch.

 

ill try to come up with a new one for you guys

[Callipygous]

i think all my favorites have already been posted on this forum : P

 

sorry.

[reverse]

I found a web page that shows some boffins solving these sorts of problems using way less obvious methods.

Now where did I put that link.

ahh here it is.

 

http://www.mathematik.uni-bielefeld.de/~sillke/PUZZLES/rope-around-earth

[reverse]

what I thought would be a good one,

was to modify this planet, rubber band one.

 

so you get the planet.

 

put a horizontal and vertical axis through it., but extend both axis right outside the circle.

 

Then overlay a square with its sides the same length as the diameter of the circle.

 

Now by rotating both shapes about their centre you will reach two interesting limits.

 

The one where the tangent is parallel to the horizontal axis, and the one where there is a 45 degree angle on each side of the ladder.

 

(You can even overlay another square, moving in exactly the mirrored direction)

 

 

 

So I guess the question is,

for what height of ladder do you calculate that your formula will first fail.

 

 

ignore what they told you. think inside the square

[Martin]

thanks for the link, reverse.

let's move on to another problem

I thought Callipygous might offer one, but he declined

 

i think all my favorites have already been posted on this forum : P

...

 

I cant remember what all has been posted and what has not!

Let's just toss out problems and see if anyone can answer. Even if it is buried somewhere probably no one will search SFN for answers, see if you can answer from your own head. If you remember the answer from an earlier SFN thread, fair game! you win (for remembering)

 

this may or may not have been asked already

 

 

continue this sequence:

 

2, 5, 5, 4, 5, 6, 3, ....

[Martin]

here's one I just thought of, maybe it is too hard, see if anything clicks:

 

this is a sequence of numbers, what sequence is it?

 

 

A, W, MG, MAL, PPPM,...

[reverse]

awwwww, did you know that your formula might fail?

it's good to know that.

[Martin]

awwwww' date=' did you know that your formula might fail?

it's good to know that.[/quote']

 

tell me where it fails, would like to know

the very next term perhaps? you tell me

 

oh shucks reverse! you seem to still be talking about the ladder!!

I thought you were talking about the sequence A, W, MG,....

 

please be explicit what you are talking about 'cause it aint always obvious

[dan19_83]

thanks for the link' date=' reverse.

let's move on to another problem

I thought Callipygous might offer one, but he declined

 

 

 

I cant remember what all has been posted and what has not!

Let's just toss out problems and see if anyone can answer. Even if it is buried somewhere probably no one will search SFN for answers, see if you can answer from your own head. If you remember the answer from an earlier SFN thread, fair game! you win (for remembering)

 

this may or may not have been asked already

 

 

continue this sequence:

 

2, 5, 5, 4, 5, 6, 3, ....[/quote']

 

 

2, 5, 5, 4, 5, 6, 3, 5, 4,

 

the first and third numbers have a difference of three, with a five in the middle. The fourth and sixth numbers have a difference of two with five in the middle. therefore the seventh and ninth will have a difference of one with a five in the middle.

i think this is wrong but i said i'd try it.

[dan19_83]

here's one I just thought of' date=' maybe it is too hard, see if anything clicks:

 

this is a sequence of numbers, what sequence is it?

 

 

A, W, MG, MAL, PPPM,...[/quote']

 

If a=1, b=2 etc. So i gave all the letters there respective numbers and added them up. didn't come up with an answer yet, just wrote it down in order to try and make i easier! Am i going in the right direction?

1, 23, 20, 26, 61, ....