Top speed?

[Andy A.]

Originally Posted by Scott Carpman

Good luck trying to turn with a 30 fps drivetrain...

Maybe this is the year that automotive style turning, with a differential, makes sense. It would certainly allow high turning speeds and easier turns for the driver to coordinate.

As to the original question, I think the general consensus is that after about 18 FPS a typical robot becomes very difficult to control and a real pig in a turn. Things like trick wheels, multiple motor drive trains, shifting gear boxes and real sharp PID loops could probably make 25 fps or higher controllable and useful, although at that speed your crossing the field very very quickly. It's an awful lot of design trade offs for that speed. But maybe thats worth it to you. If it's ever been worth it, this would be the year. I'd love to see a robot 'hotrod' that just throws every thing it has into a drive train of epic proportions.

-Andy A.

[Tottanka]

Originally Posted by Andy A.

Maybe this is the year that automotive style turning, with a differential, makes sense. It would certainly allow high turning speeds and easier turns for the driver to coordinate.

As to the original question, I think the general consensus is that after about 18 FPS a typical robot becomes very difficult to control and a real pig in a turn. Things like trick wheels, multiple motor drive trains, shifting gear boxes and real sharp PID loops could probably make 25 fps or higher controllable and useful, although at that speed your crossing the field very very quickly. It's an awful lot of design trade offs for that speed. But maybe thats worth it to you. If it's ever been worth it, this would be the year. I'd love to see a robot 'hotrod' that just throws every thing it has into a drive train of epic proportions.

-Andy A.

Did anyone say drift control?

[Gabe15]

We are definetly considering an automotive inspired steering system, but trying not to fall into making a go-cart, was more just curious for other teams sake, but I donʻt think we are going to put that much into speed...but who knows!

[MrForbes]

Originally Posted by Tottanka

Did anyone say drift control?

No, they said high speed....which is generally a different subject....

(unless you're on a dirt track)

[Protronie]

Originally Posted by Tottanka

Did anyone say drift control?

Originally Posted by squirrel

No, they said high speed....which is generally a different subject....

(unless you're on a dirt track)

Drifting robots... NOW THAT WOULD BE EXCITING!

[flamefixed]

i know this is pretty unrelated to this topic but, when coming up for new tshirt designs for this year, we decided to go with a rather known quote from a new comedy racing movie that has been out for a little while. We get to the kickoff, see the game, and WOOWWWWW!! haha.

[Daniel_LaFleur]

Originally Posted by Tottanka

Did anyone say drift control?

hehehe ... make way for sprint cars ... erm robots

Seriously though, the top speed allowed is the speed in which you can still control yout 'bot.

Anything beyond that will be 'unsafe'.

This may also be the year for suction cups on your OI, as I can see the drivers stations taking some wicked hits.

[David Brinza]

Originally Posted by Daniel_LaFleur

hehehe ... make way for sprint cars ... erm robots

Seriously though, the top speed allowed is the speed in which you can still control yout 'bot.

Anything beyond that will be 'unsafe'.

This may also be the year for suction cups on your OI, as I can see the drivers stations taking some wicked hits.

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

[Tomasz Bania]

Originally Posted by David Brinza

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

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

[663.keith]

Originally Posted by Tomasz Bania

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

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

6.4.2
Attached to the Alliance Station Wall are three aluminum shelves to support the robot control
systems of the three teams on the ALLIANCE. The support shelf measures approximately 60
inches wide by 12 inches deep. There is a 4-1/2-foot long by two-inch wide strip of Velcro tape
(“loop” side) along the center of the support shelf that may be used to secure the ROBOT controls
and Operator Interface. Each setup location includes a competition cable that attaches to the
“Competition Port” of the Operator Interface.

[Tomasz Bania]

Originally Posted by 663.keith

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

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.)

[dtengineering]

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

[WaterFreak]

Originally Posted by dtengineering

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


Jason

It sure is.......

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

[Bongle]

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.

[WaterFreak]

Originally Posted by Bongle

a perfectly efficient drivetrain and no resistance.

this is where is all goes awry, since that does not exist