We did a number of paper simulations of the average lap distance - laps with very little interference by the other robots, laps where the robot would need to weave around other robots - in an attempt to see what the average distance the robot would need to drive to score a lap. I would love to know what the CD community thinks the average distance per lap would be. Thanks!
So far, and it looks like the CD community is fairly consistant in its answers - the average lap distance is 100 - 110 feet. If so, and assuming with acceleration and deceleration - weaving through the crowds - turning, etc … how many laps can you do in the 2 minute teleoperated period? We figured, using last year’s robot, that we could do 10 ft/sec in the straight-aways, or 11 seconds with the calculated distance of the lap - BUT, with the other obstacles and turns, we figured a best guess of 16 seconds per lap, or about 8 laps per match.
How many laps do you think you could do during a match?
I also ran some calculations with this yesterday.
Using the maximum robot size (with bumpers) of ~44" x 34".
Using a CAD calculated elipse with a 870.5" (~72.54’) total length shortest travel distance per lap with the best line.
Using your given 10 ft/second robot speed…
The most we can squeeze out of this field in a perfect situation with the maximum sized robot with absolutely no traffic would be:
20 laps in 1:45
(not counting time in Hybrid mode in able to get to “the perfect path”)
Good luck with any team who can accomplish this!
I’m guessing a best lap count of about 10-15 will be very rare, and 5-10 will be the norm.
Smaller robots will have a slight advantage as they will be able to manuver around robots & the track a lot quicker, and I see rookies hopefully realizing this and taking advantage of this if they don’t want to build an arm for Ball manipulation.
I computed a version of the ‘fastest possible lap’ yesterday, and found that a lap with semicircles at each end and lines for the straights has length:
2(pir + 27 - 2r).
and that your best possible time per lap given radius r and friction coefficient u was:
t[sub]total/sub = 2pir / sqrt(9.8ur) + 4sqrt( [27-r]/[4.8u] )
Note that there are many assumptions I made in order to get ‘ideal’ lap
-Robots are points
-Robots are powerful enough that their acceleration is friction-limited
I think if you steered a 13ft radius corner at each end, thus carrying maximum speed out of each corner, a 5-6 second lap might be possible with a robot that dispenses with all other goals except speed. The best possible time for a friction-limited lap with u = 1.3 that does a 13ft corner at each end is 4.1 seconds. You can go much faster if you do a tight corner, but that’s only because we’re dealing with 1.3g acceleration along the straight.
Why 1:45?
Just curious.
-dave
Hahaha! Oops… :o
As soon as I saw your post Dave I figured there was a reason you asked that.
The explanation is that I was having flashbacks of 2 minute TOTAL time matches from yesteryear, and did my very quick calculations based on that brain hiccup and forgot the Teleoperated period time this year was 2:00 with 15 seconds of Hybrid… :ahh:
Ok… so with that said… I’ll show my work this time. lol
870.5" long path @ 120" in/sec = ~7.25 second lap time.
120 second match (Teleoperated) / ~7.25 seconds a lap = ~16.5 laps
After 15 seconds of Hybrid time, I would say a perfect line robot at maximum configuration could make in the Teleoperated mode of 2 minutes could make a total of ~16.5 laps.
Aye… my math was way off in that first attempt I guess as well. 
20 laps was a very generous overestimate I guess.
I’ll keep my prediction the same though with my original quote:
I’m guessing a best lap count of about 10-15 will be very rare, and 5-10 will be the norm.