Hello.
Our team has its turning wheels basically on a turn-table, so they both turn at the same time. Is there any formula I can use to determine turning radius, when I know how much I have ‘turned’ the turn-table off center-straight?
Hello.
Our team has its turning wheels basically on a turn-table, so they both turn at the same time. Is there any formula I can use to determine turning radius, when I know how much I have ‘turned’ the turn-table off center-straight?
I think I understand the question. Measure the angle between the axis of the two rear wheels and the axis of the two front wheels. Straight ahead the angle is zero, and if the front wheels are turned completely to one side the angle is 90 degrees.
Using that angle you can calculate the turning radius, but you need to know the distance between the front and rear wheels.
If that seems difficult, set the wheels at known angles (like 10, 20, 30 degrees) and measure the turning radius directly (draw a line on the floor and then measure the curvature)
Don
well, I have found a ‘rack and pin’ equation. But it asks for the distance between front and rear axles, and isn’t on a turntable, that ‘changing’ as I turn?
P,
If your front wheels are on a turntable so that they both turn about a central point then the turning radius is the distance between the two closest wheels. That is, when the turntable turns one of the front wheels comes closer to one of the back wheels. If you have enough travel in the turntable so that in a turn the back wheel does not move, that wheel is now the center of your turn. The radius then becomes the distance between the stationary wheel and the closest front wheel. However, your robot frame and the other wheels are still larger than that so the space you inscribe is the distance between the stationary rear wheel and the furthest point on your robot opposite that wheel. If the front wheel turntable is able to turn to a point where you could line up the center of both front wheels and that imaginary line where to intersect with the back wheel, I am guessing your robot would run into the trailer and that would become the limiting factor in your turning radius.
Simple and obvious answer: the turning radius is the distance between the robot and the center of the circle it travels while turning.
Simple observation: the wheels are moving along the circle, in a direction perpendicular to the direction to the center of the circle.
Insightful comment: The axle of each wheel is perpendicular to its direction of travel, and thus points to the center of the circle. All you need to do is find the point of intersection of the lines extending from the front and rear axles. The distance from that point to the robot is the turning radius.
Al and Alan, wouldn’t it be more correct to say that those are the minimum possible turning radii? Because steering in this case uses the slip angle of the wheels to generate turning force (I know that’s a sloppy way of wording it…) won’t the actual radius will always be greater than the theoretical minimum? On a low friction surface (like Lunacy) I would think it would be much larger than the minimum, with the radius increasing with robot speed.
thankyou for the help, are there any equations that I can get closer to ‘actual’ turning radius from the min theoretical radius?
Rick,
In cases where the robot is already moving you are correct. And in the case of a fast moving robot with the ability to have a small turnig radius (at a stopped condition) would approach infinity at it’s maximum speed. i.e. it would hit a wall or field border before turning. In cases where the robot is stationary but turning, the minimum would be the turning radius but I think it still would be affected by the trailer hitting the back of the robot.
I loved Al’s last comment so much that I think it should be spotlighted. We have these conversations all the time: “Here’s the incredibly complex calculation of how the system would work based on physics and mathematics, but the noise level is so high (or the real world will intrude in imperfect ways, like a trailer hitting a robot) that we are going to take that 27.3991 mm calculation and just turn it into ‘a big old hunk of 2-inch square tubing.’”
And this, my Padawan robot builders, is also engineering.
On the original question, doing the velocity, thrust, friction, and slip-angle math for describing turning radii at different speeds is well beyond the ability of my rusty math. Sorry.
For the theoretical aspect, just make a scale drawing with the wheels at various angles, and see where the lines intersect, as Al suggested.
For the practical aspect, put the trailer on it and drive it on FRP and see what happens. Strange things, I guarantee!
Al and Alan are correct assuming very small slip angles. This would be relatively low speed small input stuff on regolith.
Rick is really on to something. One other thing to consider is how the back wheels play into this. If they are going the same speed they will induce higher slip angle. Racers often call this Push because the back wheels are pushing the forward wheels to induce understeer (or less steer than requested). However depending on CG location and the actual manuever you attempt, you could also get Oversteer. The grossest exageration of this is the guys that do drifting, rally, or Dirt track stuff.
Gelespie (spell?) has a vehicle dynamics book that is excellent at talking about how these equations may be applied and simplified (bicycle model). Assuming reltively small slip angles, the bicycle model works well for the system you have. I.E. Take the angle off of straight. The tanget of this angle times the radius of curvature is equal to your wheelbase. Thus to find the radius of curvature:
WB=wheelbase
angle=angle off of straight that your wheels are turned.
WB/(atan(angle))=radius of curvature
If you have extra weight available, place it in different locations on your machine. This should help change the oversteer/understeer effects you might get. In general the trailer will produce understeer effects (this is what I am seeing in most videos), unless you really crack the whip, and then it could result into an oversteer condition.