Brain Exercise...

[Chris Johnson]

Seko - go back to the initial question and speak each line slowly to yourself .

[Nakranoth]

If that still fails to help... read them in pairs after line one.

[Elros]

Try reading them vertically ;- )

[Solutions Maximus]

each line describes the last

1
11 (there is one one in the last line)

[fishfin]

Everybody brobably already knows the anser to this...

There is a boat tied up to a dock with a ladder on one side. At low tide the water is at the fifth rung (counting from the top and going down). If the tide rises four feet then what rung would the water be at?

NOTE: Each rung is half a foot apart.

[Nakranoth]

The fifth rung... boat's rise with the water.

If you drop an egg from 10 high, onto a concreet floor, why will it never break?

[SekoETC]

Because "it" refers to the floor, and the egg is more fragile, thus the floor will never be the breaking party in this encounter. Am I right?

[Antichrist_Online]

You didn't define the units, thus it could be ten microns or milimetres which is a non-breaking distance if you drop it point first.

I have no riddles yet.

[Nakranoth]

Seko got it... pay attention in English class kids.

[SekoETC]

So now I should dig up a riddle... great.

Code: Select all

+----------------------->
|0             . (x,y)   +
|       ___   /|         
|    **'   '*/ |         
|  **       / **         
| *        /   |*         
|*        /   _| *       
|*       O_)__L| *       
|*               *       
| *             *         
|  **         **         
|    **-   -**           
|       '''               
V +                       


The formula for calculating the central angle in radians based on knowing values of x and y. The central point is (x0,y0).
Directions downwards and from left to right are positive. How does the radian value behave on different quadrants?

[Nakranoth]

Oh fun... you changed the layout if the quadrants... I'll let someone else answer though... I don't really feel like comeing up with any sort of riddles right now.

[SekoETC]

Code: Select all

+----------------------->
|0       |       (x,y)    NOTE: Values below are just for visualisation
|       _I_              x0=1, y0=1
|    **' | '**           
|  **    |    **         1: x=2, y= 0   2: x=0, y=0
| *   2  |   1  *        Dx= 1, Dy= -1   Dx= -1, Dy = -1
|*       |       *       
|#-------O-------#1     
|*       |       *       
| *   3  |   4  *        3: x=0, y= 2   4: x=2, y=2
|  **    |    **         
|    **- | -**           Dx= -1, Dy= 1   Dx= 1, Dy= 1
|       'T'               
V +      |               
0    1     2

Having origo in the upper corner and having the y direction reversed doesn't really have a big meaning.

Dx (delta x) = x - x0, Dy = y - x0

In 1, x > x0 and y <y0> Dx = positive, Dy = negative

In 2, x < x0 and y <y0> Dx = negative, Dy = negative

In 3, x <x0> y0. -> Dx = negative, Dy = positive

In 4, x > x0 and y > y0. -> Dx = positive, Dy = positive


tan a = Dy / Dx.

1: tan a = negative (-1)
2: tan a = positive (+1)
3: tan a = negative (-1)
4: tan a = positive (+1)

a = cot (Dy / Dx)

pi/4 = 0.7853981 rad = 45 deg
-pi/4-0.7853981 rad = -45 deg

The problem is as the angle goes over 90 degrees, it apparently doesn't care about that.

In low middle point, Dx = 0 and therefor creates an error due to division by zero.
In high middle point, Dx = 0 and Dy is also 0 so the result is 0/0 = 0.
In left middle point, Dy = 0 and Dx = -1
In right middle point, Dy = 0 and Dx = 1

Ah, screw this, I can't solve my own riddle.

Apparently pi/2, pi or 3/2 pi needs to be added depending on which quadrant is being handled. Fun.

[Nakranoth]

Come on Seko, don't remember your Trig?

Every point in the circle forms a right triangle... the angle in radians of the central angle is derived using your arcsin, arccos, and arctan functions... such that, (useing a regular coordinate system) arctan of your y value, over your x value gives you the first or fourth quadrant equivelants... in order to tell weather or not this value is correct, however, check your x value, if it is negative, you need to add pi to your answer... also, where arctan doesn't exist, you simply check your y value for positive or negative sign, and you end up with pi/4 or 3pi/4 respectively... as for an equation for the "function" it can't exist because circles have multiple y values for given x values... soo... what was I talking about?

[Pie]

The answer...

is PI!!!!

[Nakranoth]

How does the radian value behave on different quadrants?

I just realized that this was the question that was asked... I feel silly now... the radian value increases counterclockwise throught the first to fourth quadrents, then resets to zero at the end of the fourth quadrent... or keeps going, but becomes reducable to the... You know what... I don't even really care...