B Torsion Involving an Off Axis Thrust

AI Thread Summary
The discussion centers on calculating the necessary counterweight mass to maintain a vertical 2m rod during horizontal acceleration. The rod, weighing 100kg, is pushed by a car with a force applied 0.5m below its center of mass, creating a torque that requires a counterweight to prevent rotation. The counterweight's mass is derived from the relationship between the applied force and gravitational force, with calculations suggesting 5.1kg for 100N of force and 10.2kg for 200N. Participants emphasize the importance of creating a Free Body Diagram (FBD) to visualize forces and torques, noting that the moment of inertia becomes relevant only if rotation occurs. Properly accounting for all forces and their points of application is crucial for solving the problem effectively.
  • #51
Devin-M said:
The reason I ask is some electric skateboard riders showed me they are getting around 1g acceleration from leaning alone…
You can get arbitrarily high acceleration by leaning alone. Take the arc tangent of the horizontal acceleration in g's. The acceleration in g's is also the required coefficient of friction.

A drag racer can pull 5 g's. The arc tangent of 5 is around 79 degrees. If you measured the angle from the contact patch under the rear tires to the center of gravity, it would need to be under 12 degrees from the horizontal. [Ignoring aerodynamic effects]

You say that a skater can pull over 1 g. The arc tangent of 1 is 45 degrees. So the lean angle will be over 45 degrees from the vertical. You measure that angle from the contact patch under the skateboard to the position of the center of gravity of the skater plus counterweight.

If a motorcycle can corner at 2 g's then the lean angle will be 64 degrees from the vertical. You measure that angle from the contact patch to the center of gravity of rider plus bike. [Ignoring gyroscopic effects]
 
Physics news on Phys.org
  • #52
Thank you that is a very informative answer.
 
  • #53
I suggested they hold the battery forward— no weight penalty.
IMG_0317.jpeg

(AI generated)
 
  • #54
Devin-M said:
electric skateboard riders showed me they are getting around 1g acceleration from leaning alone…
I assume the forward lean (weight destitution) controls the electric motor thrust? Your phrasing is really weird. It's like saying: Fighter pilots get 9g acceleration from pulling on a stick alone.
 
  • #55
A.T. said:
I assume the forward lean (weight destitution) controls the electric motor thrust? Your phrasing is really weird. It's like saying: Fighter pilots get 9g acceleration from pulling on a stick alone.

The lean angle determines whether the rider gets “bucked off” the board when thrust is applied.

But, assuming the board is rear wheel drive, and the rider’s rear foot is above the rear truck, and the rider lifts front foot off the board during launch, then putting the battery on the rider instead of the board should give the drive wheels more traction before we even consider how it affects the angle from the contact patch to the rider’s center of mass.
 
  • #56
A.T. said:
I assume the forward lean (weight destitution) controls the electric motor thrust? Your phrasing is really weird. It's like saying: Fighter pilots get 9g acceleration from pulling on a stick alone.
Devin-M said:
The lean angle determines whether the rider gets “bucked off” the board when thrust is applied.
Does the weight distribution on the axles control the thrust?
 
  • #57
A.T. said:
Does the weight distribution on the axles control the thrust?
Yes if the rider is applying as much throttle as possible without getting bucked off.

Also yes if the drive wheel has more traction without the overall vehicle weight increasing, and the rider is applying as much throttle/thrust as possible without getting bucked off the back of the board.
 
  • #58
Devin-M said:
Yes if the rider is applying as much throttle as possible without getting bucked off.

Also yes if the drive wheel has more traction without the overall vehicle weight increasing, and the rider is applying as much throttle/thrust as possible without getting bucked off the back of the board.
How is the rider "applying throttle"?
 
  • #61
A.T. said:
If thrust is controlled by thumb, what do you mean by "getting around 1g acceleration from leaning alone"?

For the board to accelerate the rider at 1g, the center of mass of the rider needs to be leaned 45 degrees forward from where their rear foot touches the board to avoid being “bucked” off the back of the board from rotation of the body induced by the off-axis thrust (assume their front foot is lifted during launch).
 
  • #62
Devin-M said:
For the board to accelerate the rider at 1g, the center of mass of the rider needs to be leaned 45 degrees forward from where their rear foot touches the board to avoid being “bucked” off the back of the board from rotation of the body induced by the off-axis thrust (assume their front foot is lifted during launch).
Was the front foot is really lifted? How was that 45 degrees forward lean determined? For how long did that 1g acceleration persist? On what surface? What was the mass of rider + skateboard, and what was the maximal power output of the electric motor?

Note that it is trivial to measure the acceleration quite accurately with a smartphone (using its accelerometer directly, or a slow-mo camera mode and some markings on the ground). So there should better data on this than body lean angle estimations.
 
  • Like
Likes russ_watters
  • #63
A.T. said:
How was that 45 degrees forward lean determined?

I used jbriggs444’s post:

jbriggs444 said:
You say that a skater can pull over 1 g. The arc tangent of 1 is 45 degrees. So the lean angle will be over 45 degrees from the vertical. You measure that angle from the contact patch under the skateboard to the position of the center of gravity of the skater plus counterweight.

The 1g acceleration was measured with an accelerometer:

img_0197-png.png


img_0204-png.png
 
Last edited:
  • #64
Devin-M said:
The 1g acceleration was measured with an accelerometer:
View attachment 349116
View attachment 349117
You also have the speed there. Have you computed how much kinetic energy the rider + board gain during one second, and compared it to the power output of the electric motor?
 
  • #65
Well for peak power if we assume 100kg, 0m/s to 8.9m/s accelerating at 9.8m/s^2 and p = mav I get a peak power of 8.7kW, and an avg power of 1/2 the peak power or 4.3kW. Total time 0.91s and 4.3 kJ/S for total energy of about 3.9kJ.

Each of these is rated for 4kW, and you could put one on each rear wheel, but they could do a bit more power than that for a short time, either by spinning them faster with higher voltage or programming the controller to add more current for more torque:

https://flipsky.net/collections/e-s...ened-6384-190kv-4000w-for-electric-skateboard

IMG_0784.jpeg


Edit: This one rated for 5.5kw each x 2 rear wheels = 11kW

https://flipsky.net/collections/e-s...ess-dc-motor-battle-hardned-63100-190kv-5000w

IMG_0785.jpeg

IMG_0786.jpeg
 
Last edited:
  • #66
Actually if you look at those specs and calculate what’s allowed for a short time (10Nm at 10000rpm) that comes out to 10.5kW each motor or about 21kW for just the back wheels (before we even start thinking about the front wheels).
IMG_0787.jpeg
 
  • #67
Devin-M said:
Each of these is rated for 4kW, and you could put one on each rear wheel,

Edit: This one rated for 5.5kw each x 2 rear wheels = 11kW
That should indeed be enough to achieve 1g. The only remaining limit is traction: A friction coefficient of 1 is possible, but for rubber on concrete it's usually lower.
 
Back
Top