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#1
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Re: paper: Driving a Robot Fastest Path from A to B
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Here's my setup for the bigger run Start Point: (0,0) End Point: (25, -25) Max Velocity: 10 ft/s Max Acceleration: 10 ft/s^2 Max Deceleration: 15 ft/s^2 Robot Width: 2 feet Trial 1: Start Heading: 0 (facing right) End Heading: 0 Path Type: Quintic Parametric Results: Code:
Time: 4.784024461201024
Distance: 37.531625864027085
Average Speed: 7.845199406569258
Start Heading: -pi/4 (diagonal down and to the right) End Heading: - pi/4 Path Type: Line Results: Code:
Time: 4.520517763673839
Distance: 35.3518031718333
Average Speed: 7.820299580706168
![]() ![]() A great example of where a line would be much faster: ![]() Code:
Time: 4.142848102721899
Distance: 23.413771473152483
Average Speed: 5.6516123431533405
Code:
Time: 1.5516701850414147
Distance: 6.099594833091013
Average Speed: 3.9309866825392485
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#2
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Re: paper: Driving a Robot Fastest Path from A to B
Ha! Agreed!
Thanks very much for the detailed answer. Great work! One thing: The acceleration limit you used (10 ft/s/s) may be little low. Based on our speed trials, we were getting something more like 22 ft/s/s (Supershifter, hi-speed gear). I worked up a similar example using the math from my paper (I probably should have included this in the paper itself). I used a 25x25 foot path with a similarly shaped curve. I set my upper speed at 10ft/s and robot wheel base to 2 ft. Although I don't take acceleration into account, keep in mind that (a) the path is long enough that the effects of acceleration are minimized, and (b) both the linear and curved path benefit from the "instantaneous" acceleration assumption -- they are both slightly faster than real life, by about the same amount. Here are my results: - Turn-Straight-Turn: 3.69 seconds, total arc length = 35.36 feet, total heading adjustment = 90 degrees - Curve: 4.00 seconds, total arc length = 37.79, total heading adjustment = 126 degrees The curved path takes 8.3% (or 0.31 seconds) longer to execute. Although this doesn't sound like a lot, over 6 scoring cycles, that 1.84 seconds --- almost 2 seconds from optimizing just one part of a path. |
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#3
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Re: paper: Driving a Robot Fastest Path from A to B
Swerve drive teams are laughing at this thread
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#4
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Re: paper: Driving a Robot Fastest Path from A to B
I wouldn't say that. A large part of this is conservation of momentum, and the best swerve teams know that utilizing a robot's momentum to its fullest will provide the smoothest and most advantageous maneuverability the swerve design can offer. The same is true with any robot drivetrain.
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#5
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Re: paper: Driving a Robot Fastest Path from A to B
What you have brought up is a special case of a well known theorem from optimal control that says (essentially) that you always want to be saturating your inputs to get to the final state in a time optimal way. Intuitively, if your left and right motors are always going full speed in the right direction towards your setpoints (both position and angular, and you've found the right turn-straight-turn policy that ensures that this is true), you will always get there faster than if they are not going full speed all the time.
This property holds with an infinite acceleration limit and, indeed even with limits on acceleration, jerk, etc. - except in the latter case, you always want to be accelerating/jerking at the limit. The complication is that once you introduce higher order constraints, the "order of operations" policy becomes difficult to discover because of the nonlinear mapping between motor speeds and position/rotation. Once you have higher order constraints, finding a time-optimal path becomes a more difficult process, typically requiring iteration and search in a non-convex space. If you look at 254's acceleration profile generation code from this year, you'll see this concept in action - using a triple integrator with an input that always switches between +1/-1, we obtain a limited-jerk, limited-acceleration, limited-velocity trajectory. We then applied this acceleration profile over a spatial spline, which of course broke the time optimality but was good enough in practice (with some safety margin to prevent saturation). So...Yours is the right conclusion if you ignore practicalities like acceleration limits, wheel slip, battery usage, dynamics/momentum, and jerk to the robot and any load(s) it is carrying. Empirically, these are quite significant factors for FRC driving: * Robots that have high enough maximum accelerations to allow you to assume instantaneous acceleration will frequently also be capable of slipping their wheels, harming control, acceleration, and accurate distance measurement. Moreover, by the time you are talking about full-weight, 15+fps robots, the infinite acceleration assumption breaks down severely. * A robot that moves from one point to another along an arc does not need to repeatedly accelerate and decelerate the drive motors as in a turn-straight-turn case. This saves battery life, heat, and wear-and-tear over the course of a match. The battery life savings can be significant enough that it allows you to gear your robot faster overall. * Momentum is a HUGE part of FRC driving at high speeds. If I take a 254 robot at 19fps and want to turn, I can do so on a dime simply by slowing down one side of the drive and letting the momentum whip the back end of the robot around the corner. Maintaining momentum through maneuvers is generally preferential to coming to a stop. * Even if you have an infinitely accelerating robot that doesn't slip its wheels, has a perfect battery, and doesn't care about momentum, if you are carrying a game piece externally, you may want to reduce the accelerations and jerks experienced by them. Case in point: The balls we held on our bumpers during auto mode this year could be dislodged by a very abrupt stop or quick turn in place. Last edited by Jared Russell : 17-08-2014 at 12:35. |
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#6
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Re: paper: Driving a Robot Fastest Path from A to B
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Last edited by Jared : 17-08-2014 at 13:09. |
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#7
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Re: paper: Driving a Robot Fastest Path from A to B
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For 2015 and beyond, by assigning maximum velocities to each point on the path based on curvature and working backwards to respect the other limits. This is the "correct" way to do it, and fairly straightforward for the limited acceleration/unlimited jerk case. But it can be very complicated for limited jerk control. |
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#8
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Re: paper: Driving a Robot Fastest Path from A to B
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Currently, I have two approaches to the problem, neither of which seem to work very well. My first approach was to take the data for maximum velocity and find all the local minimums. These are the points in the middle of the turns where the robot should decelerate as it is approaching, and accelerate as it is driving away. Then, I would do the calculations starting at each of the minimums and work away from them in both directions, calculating velocities that could be achieved when starting at this point traveling at the maximum curve velocity, while keeping in mind acceleration limits. Then, after I had many sets of velocities, I went through and picked the smallest velocity at each point. This way, I made sure I picked a velocity that allowed me to get through every single curve without going over the speed limit. In the example graph, you can see what it would look like with a path with two minimums. The red line represents the maximum velocity the robot can travel at that point in the curve. The black line represents the velocities calculated starting from min 1 and working forward, using acceleration limits. The green line represents velocities calculated starting from min 2 and working backward, using acceleration limits. The dotted blue line is the function I'm choosing to use as velocity. Unfortunately, it has a corner at the red point where acceleration suddenly jumps, and jerk is massively large/undefined. It also takes a long time to calculate. ![]() My second approach, which is a work in progress, tries to use curve fitting (the same quintic hermite parametric spline interpolation I used for the path) to create a velocity vs. time graph. I can set the second derivatives of the start and end of the spline equal to the jerk (2nd derivative of velocity), the first derivative of the start and end of the spline to the acceleration, and the function values at the start and end of the spline to the velocity. This gives me a second derivative continuous velocity/time curve (jerk will always be finite), but jerk and acceleration may be too large. Because the splines are parametric and calculated from two nonparametric splines, I don't have set values for the derivatives and second derivatives, I only have the ratio between dx/dt and dy/dt. By manually adjusting dx/dt (and changing dy/dt to keep the ratio the same), I can change how tight the curve is. I can find a curve that meets acceleration/jerk limits, but it is time consuming and must be done manually, and I am not sure it exists for all possible paths. Last edited by Jared : 17-08-2014 at 17:19. Reason: image size |
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#9
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Re: paper: Driving a Robot Fastest Path from A to B
I realize this query wasn't directed at me, but perhaps I can contribute in a small way. Also, since I'm approaching this somewhat cold, apologies for any naοve suggestions.
Problem Statement (as I understand it): You have defined a path for the center of the robot to follow, using a parametric curve that satisfies certain end constraints (position, heading). The path may have an arbitrary number of curved segments. You wish to find control inputs (i.e., left and right motor speeds) that will cause the robot to execute that path in minimum time, while adhering to speed, acceleration and jerk constraints imposed on the motors. Is this correct? Partition the Path You can partition the path into segments between inflection points (i.e., when the curve transitions from positive to negative curvature). In the absence of acceleration and jerk limits, you would assign Vmax to either the left or right motor, correct? That is: if curvature > 0 --> Set Vright = Vmax; if curvature < 0 --> Set Vleft = Vmax; if curvature = 0, Set both to Vmax. The opposite motor will be set at whatever speed is necessary to achieve the desired curvature. So, for each path segment, you know which motor should ideally be running at max speed. Define a "Transition Zone" Based on Acceleration/Jerk Limits In reality, at each inflection point, the motors cannot transition between speed settings instantaneously because of acceleration/jerk limits. In the worst case, a motor would be commanded to go from +Vmax to -Vmax. Given the acceleration/jerk limits, it is possible to calculate the minimum arc length over which this transition can happen. Use this arc length to define transition zones on either side of each inflection point. Interpolate between Segments For each motor, create a piece-wise continuous speed curve based on segments and transition zones: There are two cases to consider: 1. Transition Zones do not overlap: In this case, for each motor join the "before" and "after" speeds with a transition curve (a piece-wise quadratic) that joins the two segments. 2. Transition Zones overlap (i.e., segment arc length is shorter than transition arc length). In such a case, the max speed you originally set for that segment is impossible to achieve. (Well, since this is based on the worst case, this isn't strictly true in all cases --- but close enough). In this case, either scale the max speed down for this segment or alter the path to give it a gentler curve. Note: This algorithm will give a sub-optimal result in terms of time, but it should be close enough. Does this help at all? |
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#10
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Re: paper: Driving a Robot Fastest Path from A to B
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Indeed, if the question had been posed as an optimal control problem, Pontryagin's Minimum Principle leads to the conclusion that "bang-bang" control inputs (where the motors are always at + or - max speed) are the optimal solution, where the cost functional is simply total time (and the upper limits of the inputs are constrained). This implies that the robot must either be in straight-line motion (both motors +) or pure rotation (one + and the other -), which corresponds to a "turn-straight-turn" approach. Any other motor state generates a curvilinear path, and would be sub-optimal (for this specific cost functional). There's a nice description here: http://www.cds.caltech.edu/~murray/b...al_04Jan10.pdf Quote:
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#11
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Re: paper: Driving a Robot Fastest Path from A to B
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#12
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Re: paper: Driving a Robot Fastest Path from A to B
Yes, the time to change heading is taken into account. The start and end headings can be arbitrary.
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#13
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Re: paper: Driving a Robot Fastest Path from A to B
I look at this and have to ask one question Jared: if you were to add PI to both of the headings, thus creating a new spline that should model a robot driving backwards, what do the numbers come out to?
Last edited by artK : 18-08-2014 at 02:22. |
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#14
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Re: paper: Driving a Robot Fastest Path from A to B
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The lower point is 8 feet away from the upper point. This takes 3.926 seconds to drive, averaging 5.86 feet per second, on a path that is 23.00 feet long. ![]() After adding pi to both headings, the path was only 7.2 feet long, took 1.573 seconds to drive, and averaged 4.57 feet per second. ![]() -Note- the acceleration counts on the top of these images are not correct, as there's some lag when using an online GIF maker, and these values are calculated in real time for the preview, just to make sure I didn't mess up calculating the values for the path file. In the actual path file, acceleration never goes below -15. Quote:
I also really like dividing based on inflection points, instead of the sharpest corner of the curve. I do see three potential issues/questions with this approach- 1. I do not always want to be accelerating/jerking at the maximum amount. If I am coming out of a tight turn, full acceleration may end up making me go too fast for the turn. Ideally, the path is straight enough that this isn't a problem, or we can just use the turn's acceleration limit in place of the robot's acceleration limit in the case of the turn's acceleration limit being lower. 2. When creating a quadratic to interpolate speeds between transitions zones, I am not sure if it can be done with just a quadratic. I need to make sure that acceleration will be a continuous function, so when interpolating, I need to be able to specify the location and first derivative of the endpoints. For this, I need at least a cubic. I also need to make sure that my acceleration and jerk never exceed certain numbers. I have no idea how I'd interpolate a curve like this, and to make it more complicated, it's more than just finding a curve whose first and second derivative are smaller than some numbers because it's a velocity vs. distance graph, so the derivatives aren't acceleration and jerk. This is the main problem I was struggling with in my second attempt. 3. When the transition zones do intersect, it becomes a very time consuming process for the computer to figure out what values will result in a slow enough turn, as it has to generate the entire path many times in a row. It is also a challenge to slow down just one part of the path, and have the robot behave following acc. and jerk limits. I can get pretty close by always picking the smallest velocity in the overlap zone (as soon as we go too fast to make the second turn, start following the second turn's velocity curve), but this ignores the jerk limit, and acceleration jumps. To be honest, I think that solving this problem with acceleration and jerk limits is beyond my ability. I think I'll just settle with ignoring the jerk limit between curves. I'll still be able to follow all the curves while keeping in mind acceleration and velocity limits, of both the robot and the curves on the path, but jerk will be forgotten at some points between tight curves. Last edited by Jared : 18-08-2014 at 09:00. |
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#15
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Re: paper: Driving a Robot Fastest Path from A to B
If you want to make sure that your path is both velocity and acceleration limited, here is a simple and widely used method:
1. For each point along the path, compute the maximum allowable velocity. For straight segments, this is simply the maximum motor velocity. For points along curves, you can compute the radius of curvature and from that derive the maximum linear speed that would let you follow the curve (based on setting the outside wheel's speed to the maximum). You now have a discontinuous velocity profile that might look something like this: Code:
Initial maximum velocities
Plot of waypoint # (x) vs. maximum velocity (y)
---- ----- --------- (fast)
------
------- (slow)
Code:
Intermediate maximum velocities
Plot of waypoint # (x) vs. maximum velocity (y)
---- ---- ------- (fast)
------/ /
-------/ (slow)
Code:
Final maximum velocities respecting acceleration limits
Plot of waypoint # (x) vs. maximum velocity (y)
--- -- ------- (fast)
\------/ \ /
\-------/ (slow)
Last edited by Jared Russell : 18-08-2014 at 14:10. Reason: distance between points, not pints... |
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