Earth is not a perfect sphere,

It's a geoid of some kind,

Not perfectly-round but close enough.

A year is a little more than 365 days.

A day is not exactly 24 hours,

little short of that.

Pi is more than 3.

Twenty-two over seven is not far off though,

often close enough.

Man hardly lives a 100 years.

Few goes beyond,

Some get close enough.

Truth ascends above digits,

distorts formulas,

bends orbits.

Man searches it with greed,

never reaches,

never conquers.

So my dear,

give up the precision,

stop running the numbers.

Instead,

smell the soil after rain,

hold a cuddling baby in your hand,

wander your vision from Vega to Orion,

let a snowflake die in your palm.

hear what morning breeze tells,

Then round up what you got.

A little more or a little less.

Even if it is still not found,

You're close enough.

--- Polar Fox

## Tuesday, February 1, 2011

### Fox 327

Labels:
Bricks,
circle,
Close enough,
Gravity,
Poems,
Rectangle,
Science's cruel - Pluto's a PLANET

Reactions: |

Subscribe to:
Post Comments (Atom)

Is this problem even solvable with the given information? Consider the base length from the leftmost weight to the rightmost weight. I believe this length varies which could change the answer to the question depending on what it was. And if the rectangle were balancing, I think the weights vary too which would change the answer. Am I just missing something here?

ReplyDeleteOh no, no. This is not a balancing problem. Not a Static problem in Physics. Assume that the brick is already fixed in the given configuration. Suppose, for example, the point it touches to the Plotu's ground holds the whole experiment firmly.

ReplyDeleteThis is a pure-geometry problem. If there is no computational errors, I don't see a problem.

-Peace

Are you sure something is not missing? Brick weight/height ratio maybe

ReplyDeleteRepeated objections made us to check the answer, we found the same result. Here, brick is a rectangle obviously, and Pluto's surface is a part of a circle. It looks like a straight line, because PLUTO IS A PLANET, after all.

ReplyDeleteIf you see a "constructional" problem, please let us know. Otherwise, this looks OK.

The answer is still classified as "not confirmed", since it has not been solved by someone else.

-8foxes

Nothing is missing. Answer is C, but I'm still looking for a geometric argument for it.

ReplyDeleteHere is the calculus approach: express each measured height to center of Pluto as 2nd degree Taylor series in 1/R (R being the radius we are looking for). You will get something like the following:

R ~ 10*20/0.00000173 cm ~ 1156 km

irrespective of brick dimensions or angle provided that R >> brick size.

bleaug

http://i51.tinypic.com/29d3r5.gif

ReplyDeleteAnd maybe I didnt understand question well

I think it's more like, consider the rightmost weight to be the top of the circle. Now since the circle barely curves near the top, the distance between the two rightmost weights are negligible (up and down). Now as you move further left on the circle, you see the same situation again, except you have a weight and a fixed point on the rectangle. Notice that this is the same distance (arc length) as the two rightmost weights. However, the up and down ratio has changed a little.

ReplyDeleteIf you just make the rightmost weight (20) the top of the circle in your picture, things should be more clear.

Like bleaug, I too am still looking for a geometric argument for it. I'm trying to imagine different angles the rectangle could be in, and something that wouldn't change given the radius of the circle. I feel like I'm close, yet I'm not quite there.

Also, the arc length isn't exactly the same; however you get the picture.

ReplyDeletelet be O the center of Pluto,r the radius,ABCD

ReplyDeletethe rectangle

OA=10.00000173+r=a,OB=30+r,OC=20+r,OD=r

the triangles OAC and OBD have the same median

hence

(r+a)²+(r+20)²=r²+(r+30)²

r=1156

http://geometri-problemleri.blogspot.com/2011/02/problem-97-ve-cozumu.html

ReplyDeleteHa, hence gravity/weights. I don't know why I didn't see that earlier; nice problem.

ReplyDeleteThank you Anonymous and Newzad for the geometric solutions.

ReplyDeleteBleaug, I really want to see the Taylor series expansion. Would you like to comment it out here?

Although, this is solved, I see merits of generating a general case for this one. Later...

well if you insist here it is:

ReplyDeletehttp://bleaug.free.fr/8foxes/8foxes327.pdf

I apologize in advance for the overpowered, tedious and lengthy calculations contained in that document compared the one-line elegant geometric argument given by Anonymous above :-D

Bleaug

A solution is a solution. It may be useful for those who study calculus.

ReplyDeleteThank you Bleaug.