Torsion Involving an Off Axis Thrust

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SUMMARY

The discussion centers on calculating the necessary counterweight mass for a 2m, 100kg uniform rod being accelerated horizontally while remaining vertical. The force is applied 0.5m below the center of mass, creating a torque that requires a counterweight to prevent rotation. The equations derived include mc = F / g and mc = a * mt / g, where mc is the counterweight mass, F is the force applied, g is the acceleration due to gravity, and mt is the total mass. The calculations show that for a 100N force, a counterweight of 5.1kg is needed for 1m/s² acceleration, and 10.2kg for 2m/s² acceleration.

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  • #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).
 
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  • #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.
 
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  • #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
 
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  • #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
 
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  • #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.
 

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