***Assume the following:
skid-steer vehicle with 4 standard non-treaded wheels
square drivebase: trackwidth = 2 ft, wheelbase = 2ft
coefficient of kinetic friction 0.7, independent of speed and direction
center of mass 150 pounds is located at center of geometry
left wheels being driven forward at tangential speed 2 ft/sec
right wheels being driven backward at tangential speed 2 ft/sec
At steady state, how fast is the robot rotating?**
*
I’m not sure how friction factors into this. A really wide wheel will have much more turning scrub than a thin or donut shaped wheel.
Here’s what I’ve got so far:
I broke the wheel velocity into two components, one in the direction from the wheel to the center of the bot, and the other perpendicular to this one. The inward pointing one is all lost to friction. The other one has a magnitude of 0.707 * 2 = 1.414 feet/second.
The distance from the center to the wheel is 2.828 feet. This means one rotation is 2pir=17.768 feet. This means it’s rotating at 1.414/17.768 = 0.079 rotations per second.
I think I’m missing something eight the friction.
I’m not getting the friction part yet either, though I think you may have miscalculated the radius you used, 2.828 is the distance from wheel to wheel diagonally, making the rotation .159 rev/s. With the friction I’m guessing it will be less than this.
I would think the coefficient of kinetic friction (and robot mass, for that matter) comes into play when calculating torque required to continue to rotate at the determined steady-state speed. But this was not asked.
My first thought was 2rad/sec, but I’m pretty sure that’s with 4 omni wheels. In order to solve this you need to know how much linear distance is traveled, when 2ft of tire tread is pulled by.
Oops. I change my answer to .158 rotations per second.
Just where do you define the tangential speed? Tangent to the circle subscribed through the wheels center line? if so rotation is 2ft/s* C/8.884ft=.2251rev/sec.
If “Tangential speed” is the slip between the wheel & the ground normal to the wheel axis, then life becomes more complicated & you need weight & friction. For instance a coefficient of 0 rotational speed is uncoupled from wheel speed.
Reps to you Nate.
Given the assumptions in the problem statement, the answer is 1 rad/sec.
The wheel radius times the wheel radians/sec, tangent to the circumference of the wheel at the contact patch, in the plane of the wheel.
At steady-state, the net torque acting on the bot around its center of mass must be ZERO.
The only forces acting on the bot in the horizontal plane are the four friction forces (one at each wheel). The direction of these forces will start out aligned with the plane of each wheel, but will change as the robot begins to spin. Given the assumption about CoF being independent of speed and direction, the friction at each wheel will always point in the direction of the relative motion between the carpet and the wheel at the contact patch. This relative motion is the vector sum of the wheel tangential speed plus the carpet speed at the wheel due to the robot’s motion.
When the robot spin reaches the point where the net torque around the center of rotation is zero, the spin will stop changing.
The robot weight (150 lbs) and CoF (0.8) were red herrings. Given the assumptions, their values do not affect the answer (as long as neither is zero).
*Physics Quiz 10 part B
Now, assume the drivebase is rectangular instead of square:
wheelbase = 2 feet, trackwidth = (28/12) feet. Everything else the same.
At steady state, how fast is the robot rotating?
.85 rad/s
The rotation speed will be achieved when the net torque on the robot becomes zero (after it was non-zero while the robot is initially accelerating). Also, there is only one force coming from each wheel, the frictional force provided by the carpet on the wheels. Therefore, the only way bring this force to zero (since it has a non-zero magnitude) is to have its line of action be in line with the center of rotation. (such that cos(theta) = cos(180) = 0 in the torque calculation).
The frictional force will also be in line with the vector sum of the wheel’s tangential speed and the carpets speed.
Knowing the tangential speed of the wheel, the tangential direction of the robot’s spin and the direction of the frictional force at the wanted equilibrium, the speed of the robot’s spin can be solved for.
Calculations
[spoiler]Vcarpet = Vc = opposite Vrobot
F = friction (not the actual force but friction lies along this vector, this is technically the vector sum mentioned earlier)
2 + Vcy = Fy
Vcx = Fx
Fy/Fx = -(28/12) / 2 #Known angle of vector sum/friction
Vcx/Vcy = (28/12) / 2 #Known angle of carpet velocity
Vcy = -.847
Vcx = -.988
Vc magnitude = Vrobot magnitude = 1.302 ft/s
Vrobot = r * omega
omega = 1.302 / (sqrt(2^2 + (28/12)^2) / 2) = .847 rad/s
crosses fingers
[/spoiler]
What formula did you use to calculate the above? Show your work please.
omega = 1.302 / (sqrt(2^2 + (28/12)^2) / 2) = .847 rad/s
Does that value of omega agree with your Vcy and Vcx calculations above??
cos(tan^-1(12/14) = 0.7593
0.7593*2 = 1.5186 ft/second
Length of diagonal is 3.0731 feet.
Circumference is 3.0731*pi = 9.6544
1.5186/9.6544 =0.1572 rotation/sec = around 1 rad/sec.
Its a system of 4 equations, I just solved for Vcy and and then got Vcx from the ratio of Vcy/Vcx.
Solving
[spoiler]2 + Vcy = Fy
Vcx = Fx
Fy/Fx = -(28/12) / 2 #Known angle of vector sum/friction
Vcx/Vcy = (28/12) / 2 #Known angle of carpet velocity
Substitute in for Vcx and Fx:
2 + Vcy = Fy
((28/12)/2) * Vcy = (-2/(28/12)) * Fy
Substitute in for Fy:
2 + Vcy = ((-(28/12)^2)/(2^2)) * Vcy
Vcy = -.847
Into another equation:
Vcx/(-.847) = (28/12)/2
Vcx = -.988
Robot Velocity components are opposite of these.
Vry = .847
Vrx = .988
Vr = sqrt(.847^2 + .988^2) = 1.302 ft/s
Vtangential = r * omega
r = d/2 = sqrt(2^2 + (28/12)^2)/2 = 1.537 ft
omega = Vt / r = 1.302 / 1.537 = .847 rad/s
[/spoiler]
Yes, via the equation Vt = r * omega
Take your value omega=0.847 rad/sec and use it to calculate the x and y components of carpet speed:
Vcx = omega*(L/2) = 0.847*(2/2) = 0.847 ≠ 0.988
Vcy = 2-omega*(W/2) = 2-0.847*((28/12)/2) = 1.012 ≠ 0.847
Now try omega = 84/85:
Vcx = omega*(L/2) = (84/85)*(2/2) = 0.988
Vcy = 2-omega*(W/2) = 2-(84/85)*((28/12)/2) = 0.847
In this particular problem, you can get the correct answer – as you have done – by simply projecting the wheel speed unto the direction the wheel is moving.
But as the problem gets more general (and more complicated), that will no longer work.
Oh, it would seem that I made a stupid error. I swapped what I used for Wheelbase and Trackwidth. I wasnt sure, so I looked it up online and people were saying that trackwidth is the distance between wheels on a vehicle. It seems thats not what you used here.
I swapped my values in the calculations and my method gave me the correct answer. When I get home I’ll repost a correct explanation.
Yes, trackwidth is the distance between the 2 front wheels. (Or the 2 rear wheels). Here’s an easy way to remember: It’s the width of the “track” left by the vehicle when traveling forward in a straight line.
It seems thats not what you used here.
That is what I used.
Ah, well even if I did find out the dimensions correctly I must have set up the proportion between the carpets velocity and the robot’s orientation inversely. The solution below works out, so unless a mathematical equality that makes this give the right answer for the wrong reason is working here and I haven’t realized what it is, I must have accidentally flipped that value somehow. It was late, I probably used the wrong angle for being tangential to the robots rotation which lead me to the wrong proportion.
Updated solution:
Solution
[spoiler]
2 + Vcy = Fy
Vcx = Fx
Fy/Fx = -2 / (28/12) #Known angle of vector sum/friction
Vcx/Vcy = 2 / (28/12) #Known angle of carpet velocity
Substitute in for Vcx and Fx:
2 + Vcy = Fy
(2 / (28/12)) * Vcy = (-(28/12) / 2) * Fy
Substitute in for Fy:
2 + Vcy = ((-2^2)/((28/12)^2)) * Vcy
Vcy = -1.153
Into another equation:
Vcx/(-1.153) = 2 / (28/12)
Vcx = -.988
Robot Velocity components are opposite of these.
Vry = 1.153
Vrx = .988
Vr = sqrt(1.153^2 + .988^2) = 1.519 ft/s
Vtangential = r * omega
r = d/2 = sqrt(2^2 + (28/12)^2)/2 = 1.537 ft
omega = Vt / r = 1.519 / 1.537 = 84/85 = .988 rad/s
I confirmed this using the other method and my answer was the same down to 6 decimal places.
[/spoiler]
