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asd
#17

anon
Reply 0 123456789123345869
(02/19/2013) [] shouldn't the square's graph lack a point of inflection when y is changing? If the graph is the y coordinate of the shape over a constantly increasing third parameter, I'm pretty sure the square's graph should be straight lines.
#18 to #17

anon
Reply 0 123456789123345869
(02/19/2013) [] Ahh ok. Got it. The third parameter's rate of change isn't constant. The graph is organized so that the angle of the line on the shapes are uniform. Since the square has a larger perimeter, the exterior point has to move faster at certain spots so the angle is the same with the circle so it looks pretty.
#53 to #17

popkornking
Reply 0 123456789123345869
(02/20/2013) [] Oh look at that, anon's name changed again...
#66 to #17

JonathanNowFuckYou
Reply 0 123456789123345869
(02/20/2013) [] Well, if we assume that the line is just spinning around the center of the shape, the dot on the out side of the square fluctuates in speed as the line gets further away from the center and yet has to maintain the constant speed.
#255 to #17

younglegend
Reply 0 123456789123345869
(02/20/2013) [] It would be a straight line if the graph wasn't moving
#118 to #17

toolbox If the point going around the perimeter of the square traveled at a constant speed then yes. The slop would be 1 to 1 or 2 to 1 or whatever, but it would graph a straight line. However, here the constant is the angular velocity, and (as previously stated) since the points at the midpoint of the square section are closer than the points at the corners, the speed of the 'point' following the perimeter increases as the distance from the center of the circle increases.
Think of a door on a hinge (as usual). If you are opening a door and you consider a point 1 inch away from the hinge, the angular velocity of that point is considerably smaller than if you consider a point at the end of the door (relative to the hinge), which is about 3 feet away.
Reply +3 123456789123345869
(02/20/2013) [] Think of a door on a hinge (as usual). If you are opening a door and you consider a point 1 inch away from the hinge, the angular velocity of that point is considerably smaller than if you consider a point at the end of the door (relative to the hinge), which is about 3 feet away.
#290 to #118

treska
Reply 0 123456789123345869
(02/20/2013) [] Your comment is very close to a perfect explanation, but i think that when you wrote the example of the door, your mind went confused. I think that where you wrote "...,the angular velocity of that point..." you meant the** linear velocity**(or whatever you say in english, i'm spanish), didn't you? That's since you are considering a door opening with constant angular velocity.
#196 to #17

thatguynobodylikes **User deleted account**
+3 123456789123345869
has deleted their comment [] #83 to #1

TheFunnyJunkie Look at the left girl then to the middle, it matches.
Look at the right girl then to the middle, it matches....
this one ***** with my mind.
Reply +8 123456789123345869
(02/20/2013) [] Look at the right girl then to the middle, it matches....
this one ***** with my mind.
#72 to #57

swimmingprodigy you need to find yourself a girl, mate
Reply 2 123456789123345869
(02/20/2013) [] #88 to #78

swimmingprodigy
Reply 12 123456789123345869
(02/20/2013) [] yes but there's a difference between "basic understanding of algebra/geometry" and writing out a proof for a website that supposed to contain nothing but funny pictures
#74 to #57

rubberhose i'm a big fan of tangent functions. i once spent my entire walk to school figuring out the length of a shadow as a function of the time of day.
Reply 0 123456789123345869
(02/20/2013) [] #150 to #57

partnerintroll
Reply +1 123456789123345869
(02/20/2013) [] In layman's terms if you've passed PreAlgebra:
Treat the line from the center to the outer edge as an angle attached to another line segment that runs from the center to the right in a perpendicular way. The graph to the right is the function value for whatever angle is formed by the rotating line and the fixed one, and creates the waves shown
Treat the line from the center to the outer edge as an angle attached to another line segment that runs from the center to the right in a perpendicular way. The graph to the right is the function value for whatever angle is formed by the rotating line and the fixed one, and creates the waves shown
#139

ithinkimfunny Fouriertransforms up in this ************
Reply +39 123456789123345869
(02/20/2013) [] #262 to #160

ponieskilledmyacc
Reply 0 123456789123345869
(02/20/2013) [] sadly enough, my teacher told all of my class to use a little rope to measure the circumference and the diameter of a couple of circles then divide one by the other, before teaching us about pi. I was the only one in the class that didnt get ~3.1415...
#65 to #28

voidekatwo
Reply 0 123456789123345869
(02/20/2013) [] Do you know the name of this because it looks incredibly useful o.o
#218 to #65

evilpapagali
Reply +1 123456789123345869
(02/20/2013) [] It is called malta cross drive or geneva mechanism.
#228 to #30

voidekatwo I think this one is pretty cool, it reduces the incoming rotational speed by a significant amount :P
Reply +1 123456789123345869
(02/20/2013) [] #125 to #29

rockamekishiko
Reply 3 123456789123345869
(02/20/2013) [] is that how airplanes rotors work?
#120 to #100

randomnezz
Reply 0 123456789123345869
(02/20/2013) [] Hate to be an arsehole here (Yes, I'm Australian, get over it)
But the top graph can either be sine or cosine. As they're effectively the same except one is shifting half a period along.
The second somewhat displays a tan function (Apart from the straight parts of the graph)
The third graph doesn't show any actual equation...
TL:DR A rant about the graphs from a mathematics major from Australia.
Inb4 ********* of red thumbs.
But the top graph can either be sine or cosine. As they're effectively the same except one is shifting half a period along.
The second somewhat displays a tan function (Apart from the straight parts of the graph)
The third graph doesn't show any actual equation...
TL:DR A rant about the graphs from a mathematics major from Australia.
Inb4 ********* of red thumbs.
#54

scotfighterz
Reply 13 123456789123345869
(02/20/2013) [] > 3 graphs
> year 201"3"
> almost march which is the 3rd month
> HALO 3 confirmed
> year 201"3"
> almost march which is the 3rd month
> HALO 3 confirmed
#99 to #54

stainunderyournose
Reply 4 123456789123345869
(02/20/2013) [] should have said halflife 3, or portal 3, or even team fortress 3!
#67 to #62

ryderjamesbudde **User deleted account**
0 123456789123345869
has deleted their comment [] #84 to #58

scotfighterz
Reply 1 123456789123345869
(02/20/2013) [] Yeah I should probably just not even bother with the ____ 3 jokes anymore, since I suck terribly at them