How does descoring work?

[Ragnarokae]

So you could go past the overpass, then turn around and knock the ball off...

Besides, G22 says "by moving", not by turning.

<G22> Direction Of Traffic – ROBOTS must proceed around the TRACK in a counter-clockwise direction. Once a ROBOT has CROSSED a LANE MARKER or FINISH LINE, it shall not break the plane of the line by moving in the clockwise direction. A PENALTY will be assigned for each infraction.

[Ragnarokae]

I agree w/ the ball hanging idea, any way to calculate how low it hangs?

[dbell]

Originally Posted by Ragnarokae

I agree w/ the ball hanging idea, any way to calculate how low it hangs?

We measured about 70 inches from the bottom of the ball the floor, at the kickoff.

[Ragnarokae]

Originally Posted by dbell

We measured about 70 inches from the bottom of the ball the floor, at the kickoff.

Is that measurement, or were you able to calculate that? If it was calculation, I would really like to know how for future reference.

[Bongle]

Originally Posted by Ragnarokae

Is that measurement, or were you able to calculate that? If it was calculation, I would really like to know how for future reference.

Posted this in a couple other threads today.

given a trackball center at (0,0)
given a trackball contact point with a rail at (16,y) (the rails are 32" apart)
given the trackball is 20" and assuming it remains approximately spherical, then...

x^2 + y^2 = r^2
16^2 + y^2 = 400
y^2 = 400 - 256
y = 12

So the contact point for the rails will be 12" below the center of the ball. This means that between the rails, the bottom of the ball should be 8" below them, and thus 2" below the maximum allowed height of a robot in the opponent zone. However, this is in an ideal world where the rails are infinitely thin and the ball is perfectly sperical. Since the rails are 1.5" wide and the ball will deform some, the ball may be slightly higher than where I computed it to be, and the contact space for a robot will be very, very small.

Glad to hear someone did the actual measurement and it was close

Here's a diagram:
http://www.chiefdelphi.com/forums/at...4&d=1199558995

[GBIT]

given a trackball center at (0,0)
given a trackball contact point with a rail at (16,y) (the rails are 32" apart)
given the trackball is 20" and assuming it remains approximately spherical, then...

x^2 + y^2 = r^2
16^2 + y^2 = 400
y^2 = 400 - 256
y = 12

So the contact point for the rails will be 12" below the center of the ball. This means that between the rails, the bottom of the ball should be 8" below them, and thus 2" below the maximum allowed height of a robot in the opponent zone. However, this is in an ideal world where the rails are infinitely thin and the ball is perfectly sperical. Since the rails are 1.5" wide and the ball will deform some, the ball may be slightly higher than where I computed it to be, and the contact space for a robot will be very, very small.

Originally Posted by Bongle

Posted this in a couple other threads today.


Glad to hear someone did the actual measurement and it was close

Here's a diagram:
http://www.chiefdelphi.com/forums/at...4&d=1199558995

Umm. Hope not too many people used this number because our team found it to be wrong. We noticed the the equation above used numbers assuming the ball was on center with the pipes. this is not true seeing that the ball rests on the top of the pipes. We found the hang to be (assuming the the 6.5' overpass is on center with the pipes) it would hang only 1.5 inches below the 6' legal limit. If the 6.5' is at the bottom of the pipes it would hang only .67 inches from the 6' limit. this is assuming the the ball in of 40" diameter.



here is the diagram:
http://www.chiefdelphi.com/forums/at...6&d=1199655203

[dbell]

Originally Posted by Ragnarokae

Is that measurement, or were you able to calculate that? If it was calculation, I would really like to know how for future reference.

We measured that at the kickoff.

But it is possible to figure it out. The overpass bars are 32 inches apart.(the arena 6.2.2) The balls are 40" in diameter.(the game piece 6.3) And the overpass is 78 inches high.
I'll let you do the math.

[LordTalps]

Originally Posted by Ragnarokae

Is that measurement, or were you able to calculate that? If it was calculation, I would really like to know how for future reference.

Circular relationship, err, function for a circle. X^2 + Y^2 = R^2

Radius of ball -- 20"
length of overpass -- 32", believedly.

16^2 + y^2 = 20^2
y = 12. 12 is the height from the center of the sphere to the overpass.

20 - 12 = 8" the ball dips down at maximum, giving us only 2" of impact room to play with.

edit: b10. bleh.