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-   -   Physics Quiz 10 (http://www.chiefdelphi.com/forums/showthread.php?t=131246)

Ether 26-11-2014 11:01
Re: Physics Quiz 10
 
Quote:
Originally Posted by dellagd (Post 1410148)
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

Ether 26-11-2014 11:05
Re: Physics Quiz 10
 
Quote:
Originally Posted by Jared (Post 1410146)
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.
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.

dellagd 26-11-2014 11:42
Re: Physics Quiz 10
 
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.

Ether 26-11-2014 11:57
Re: Physics Quiz 10
 
Quote:
Originally Posted by dellagd (Post 1410167)
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.
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.

Quote:
It seems thats not what you used here.
That is what I used.

dellagd 26-11-2014 12:00
Re: Physics Quiz 10
 
Quote:
Originally Posted by Ether (Post 1410177)
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.
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:
Spoiler for Solution:

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.

Ether 26-11-2014 19:00
Re: Physics Quiz 10
 

Physics Quiz 10 part C:


Given:

- wheelbase=24 inches, trackwidth=30 inches

- center of mass 150lb located 3 inches forward and 4 inches to the right of the center of geometry

- assume flat level floor, and the bottom of all four wheels coplanar


Find the weight (normal force) at each wheel.

Jared 26-11-2014 19:34
Re: Physics Quiz 10
 
Quote:
Originally Posted by Ether (Post 1410236)

Physics Quiz 10 part C:


Given:

- wheelbase=24 inches, trackwidth=30 inches

- center of mass 150lb located 3 inches forward and 4 inches to the right of the center of geometry

- assume flat level floor, and the bottom of all four wheels coplanar


Find the weight (normal force) at each wheel.
Front left 34.375
Front right 59.375
Back left 20.625
Back right 35.625

Work:
http://i.imgur.com/MQ0xHY7.jpg

Ether 26-11-2014 19:37
Re: Physics Quiz 10
 
Quote:
Originally Posted by Jared (Post 1410242)
Front left 34.375
Front right 59.375
Back left 20.625
Back right 35.625

Work:
Reps to you !

Ether 26-11-2014 20:03
Re: Physics Quiz 10
 

Physics Quiz 10 part D:


Assume the following:

- wheelbase=24 inches, trackwidth=30 inches

- center of mass 150lb located 3 inches forward and 4 inches to the right of the center of geometry

- assume flat level floor, and the bottom of all four wheels coplanar

- skid-steer vehicle with 4 standard non-treaded wheels, except the 2 front wheels are badly misaligned:
--- FrontRight is 0.22983514251446 radians CCW
--- FrontLeft is 0.65426122466113 radians CW

- coefficient of kinetic friction 0.7, independent of speed and direction

- 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, the vehicle's center of rotation is located at the center of mass

At steady state, how fast is the robot rotating?

Jared 26-11-2014 20:50
Re: Physics Quiz 10
 
This one is trickier...

Im not sure if this is right, especially because the robot must now be constrained to rotate around its center, but I got 0.1748 rotations per second = 1.098 rad/sec.

Work:

Ether 26-11-2014 21:28
Re: Physics Quiz 10
 
Quote:
Originally Posted by Jared (Post 1410250)
This one is trickier...
Yes it is. But this is all leading somewhere, and we're almost there.

Quote:
Im not sure if this is right, especially because the robot must now be constrained to rotate around its center, but I got 0.1748 rotations per second = 1.098 rad/sec.

Work:
There was no link to your work.

Here's how you can check your answer:

1) at each wheel, use your answer to calculate the carpet velocity vector at that wheel due to robot rotation around the center of mass

2) at each wheel, combine (vector addition) the carpet velocity due to robot motion at that wheel (from step1) and the tangential velocity of that wheel to get the net carpet velocity at that wheel.

3) compute the normal force at each wheel, and use that to calculate the magnitude of the kinetic friction at each wheel.

4) at each wheel, use the net carpet velocity direction (from step2) at that wheel to split the kinetic friction magnitude at that wheel into X and Y components at that wheel

5) Sum the X components from step4. The sum should be zero at steady state.

6) Sum the Y components from step4. The sum should be zero at steady state.

7) Sum the torques around the center of mass due to the X and Y components from step4. The sum should be zero at steady state.

8) If any of the sums from steps 5, 6, or 7 are not zero, your answer is not correct.

Jared 26-11-2014 21:43
Re: Physics Quiz 10
 
I think now that my answer is wrong because I must have a net force in the x direction due to friction. For my answer I had used my previous method of breaking the velocity into two perpendicular components, with one of those components tangent to the circle centered at the robots center and intersecting the wheel.
I then multiplied each of these velocities by the ratio of weight over the wheel to total weight so that the wheel closer to the center of mass and with the most weight on top of it have a greater impact on total velocity.

To me, this seems like you might be leading Us toward forward kinematics for swerve...

Ether 27-11-2014 10:39
Re: Physics Quiz 10
 
Quote:
Originally Posted by Ether (Post 1410253)
Here's how you can check your answer:
Notice that you can use the above to find the answer.

Set up equations using steps 1 thru 7, then plot 5, 6, and 7 vs omega and locate the value of omega for which all 3 curves are zero.

Reps to the first person to post a graph with the answer (show your work).

Ether 30-11-2014 15:06
Re: Physics Quiz 10
 
Quote:
Originally Posted by Ether (Post 1410253)
Here's how you can check your answer...
This thread keeps getting hits so maybe there's still some interest.

I'll kick-start this, with Step#1 for the Front Right wheel:

Code:


> # rad/sec robot rotation around CoM:
> omega=1.098
      1.098
>
> # wheelbase:
> L=24/12
          2
>
> # trackwidth:
> W=30/12
        2.5
>
> # X coordinate of Center of Mass (relative to Center of Geometry):
> Xm=4/12
 0.33333333
>
> # Y coordinate of Center of Mass (relative to Center of Geometry):
> Ym=3/12
      0.25
>
> # X component of carpet motion at FrontRight wheel due to robot rotation:
> V1x=-omega*(L/2-Ym)
    -0.8235
>
> # Y component of carpet motion at FrontRight wheel due to robot rotation:
> V1y=omega*(W/2-Xm)
    1.0065

Ether 04-12-2014 17:29
Re: Physics Quiz 10
 
1 Attachment(s)

Steps 1 & 2 for front right wheel.


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