Re: Top speed?
 

There are long Velcro strips provided at the player stations this year - guess the GDC saw this coming!!

Re: Top speed?
 

But AFAIK, they are right in the middle of the stations therefore i see some teams having control mounting conflicts.

Re: Top speed?
 

the Velcro is clearly described in the rulebook, so teams should incorporate this into the design of their operator interface for this year.

Re: Top speed?
 

Agreed, and it's a good thing the GDC thought of this. Even the 7-10lb trackball could possibly knock down an unsecured interface, and this will be a guaranteed equipment saver during the hybrid period (Read as: 120LB robot plows into the alliance station at 12 ft/sec with no one holding the controls.)

Re: Top speed?
 

The laws of physics actually impose much lower speed limits than c, or even Mach 1.

Consider that the robot has to navigate a corner at each end of the track. A high traction wheel will have a co-efficient of friction of about 1.3. This limits the speed in the corner... or at least the speed that can be maintained through the corner without skidding into the walls.

It also limits the accelleration of the robot down the straights, and requires the driver to slow the robot before entering a corner (assuming that straight speed exceeds the maximum cornering speed.)

Some fairly simple kinematics calculations should allow teams to calculate the theoretical top speed and fastest lap time possible for a robot.

By the way, I note Squirell's reference to NASCAR... but isn't this more like NASACAR?


Jason

Re: Top speed?
 

They had them last year too. I'm encouraging my team to put Velcro on the bottom of their board.

Re: Top speed?
 

i think to fast could do damage to your own robot if not others lol try taking the 90 degree turns really fast and tell me you dont hit the wall

Re: Top speed?
 

A low center of gravity would help, although you COULD spin out as a result...maybe 20 ft/sec drive train is a little too fast?

Re: Top speed?
 

I think this year there will be robots exceeding 20 fps but these robots will also probably has sophisticated turning systems or eliminate the need to turn entirely (20 fps mecanum/holonomic/swerve). I think teams that used swerves last year may have an advantage the image of team 71 swerving around the rack at high speed sticks in my mind lol.

Re: Top speed?
 

Does anyone know if there's any rule about an unsafe speed or is that just a judge related matter?

Re: Top speed?
 

While I agree that you will see many robots going faster than 20fps, I doubt very many teams will do well that go that fast. The reason I say that is because winning races has more to do with control than speed, especially on a short track such as FIRST has provided.

No. I believe that this year the winners will be the robots with the best control over their drive system and the robots that can take the big hits from those that cannot control their drive systems.

Re: Top speed?
 

Seems like a decent way to take the turns...:p

Re: Top speed?
 

I scrolled through the manual, but I didn't see a specific rule.
Generally, the refs judge "unsafe speed" based on whether or not you have control over your machine. If you're going flying into walls, if you keep heading toward the outside of the field, if you're ramming into other robots, be prepared to be deemed unsafe.

But as long as you seem to VISIBLY have a handle on your machine and what it's doing, then you should be fine.

Re: Top speed?
 

It sure is.......

Go straight, turn left, turn left. Repeat endlessly!!!!!!!!! :D :D :D

Re: Top speed?
 

Alright, I'm doing the computations for the fastest theoretical lap.

I'm saying the straight is 27ft = 8.22m long.

A friction-limited single lap of overdrive includes:
A) a half-circle of radius r and length pi*r
B) a straight with [(8.22- 2r)/2]m of acceleration followed by (8.22 - 2r)/2]m of deceleration for a total of (8.22-2r) distance
A) another half-circle of radius r and length pi*r
B) another straight


Total distance: 2*pi*r + 2(8.22-2r) = 2*pi*r + 16.44 - 4r

Time tc for the circles will be determined by the centripetal equations.
t_c = pi*r / v_c
F_c = Ff = uFn = umg = -9.8mu
-9.8mu = -mv_cv_c/r
9.8u = v_cv_c/r
9.8ur = v_cv_c
v_c = sqrt(9.8ur)
t_c = pi*r / sqrt(9.8ur)

Time ts for the straights will be determined by straight-line acceleration and deceleration, which means we need the kinematic equations. Starting and ending speed for an ideal lap will be vc (our exit speed from the corners). Our acceleration will be friction-limited. Our distance is half the straight, which is (8.22-2r)/2 or 4.11-r
F = ma
Ff = ufn = 9.8mu
9.8mu = ma
9.8u = a // this is our acceleration due to friction


d = v_c*t + 0.5*a*t_s*t_s
4.11-r = sqrt(9.8ur)*0 + 0.5*(9.8u)*t_s*t_s
4.11-r = 4.8*u*t_s*t_s
sqrt( [4.11-r]/[4.8u] ) = t_s

So now that we know our time for the half-circles (t_c) and straights (2t_s), we can add it all together. There are 2 half-circles, 2 acceleration phases and 2 deceleration phases in a single friction-limited lap.
t_total = 2t_c + 4t_s
t_total = 2*pi*r / sqrt(9.8ur) + 4*sqrt( [4.11-r]/[4.8u] )

If you like calculus, sub in a known friction coefficient for u and then solve for the optimum turning radius. Since I have long since forgotten how to do a complicated derivative like I'd need to do here, I'm just going to sub in r = 13ft = 3.96m and u = 1.3.

t_total = 4.12 sec (laps per minute: 14)
Cornering speed = vc = sqrt(9.8ur) = 7.1m/s = 23fps

Edit: After graphing the result for u=1.3, I see that a 3.96m radius turn is very close to the WORST you can do for a friction-limited lap. You could be doing 2 second laps if you had a 50cm turn radius. However, the power output required to accelerate a 50kg robot at 1.3gs and slow it back down twice per lap is ludicrously high. Specifically, if you accelerated a 50kg robot at 1.3g for the 4.11m and decelerated just as fast for 4.11m, then you're looking at this much work:
W = F*d
W = 1.3*9.8*50*8.22 = 5236J of energy expended

Since we know our time for the 8.22m straights is 2*t_s = 2*sqrt( [4.11]/[4.8u] ) = 1.62s for the whole straight. This means your power output will be 3.2kW, which at 12 volts means you're drawing 268amps on the straights, assuming a perfectly efficient drivetrain and no resistance.