How Much Torque is Needed to Overcome an Obstacle with a Rotating Wheel?

AI Thread Summary
To determine the torque needed for a wheel to overcome an obstacle, one must consider the wheel's mass, radius, and the height of the obstacle. The energy required to clear the obstacle can be calculated using the formula E=mgh, where 'm' is the mass, 'g' is gravitational acceleration, and 'h' is the obstacle height. A larger wheel radius allows for easier passage over obstacles due to better leverage and balance. The force required to push the wheel over the obstacle can be expressed as F=m*g*sqrt(2rh-h^2)/(r-h), and the torque can be calculated using T=F*r. Understanding these relationships is crucial for calculating the necessary torque and power for a rotating wheel to maintain speed while overcoming obstacles.
vladim1
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Hi to all

Please help me to clarify this situation:

There is a rotated wheel on a plan surface. What torque/power will be necessary to spring an obstacle with a concrete height.

For ex. here are some data:

1.Wheel mass 10 kg
2.Wheel radius 0,3 m
3.Rotation speed 100 RPM
4.Obstacle height 0,01 m

Which formulas must I use to determine torque/power to get the wheel rotating at that speed without obstacle, and how much must be increased these parameters to pass obstacle without slowly speed.

Here is an image, hope it will help to understand what I mean.

Thanks
 

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vladim1 said:
Hi to all

Please help me to clarify this situation:

There is a rotated wheel on a plan surface. What torque/power will be necessary to spring an obstacle with a concrete height.

For ex. here are some data:

1.Wheel mass 10 kg
2.Wheel radius 0,3 m
3.Rotation speed 100 RPM
4.Obstacle height 0,01 m

Which formulas must I use to determine torque/power to get the wheel rotating at that speed without obstacle, and how much must be increased these parameters to pass obstacle without slowly speed.

Here is an image, hope it will help to understand what I mean.

Thanks

Hi vladim1! Welcome to PF! :smile:

Isn't this just an energy problem?

As son as the roller hits the obstacle, it will start rotating about the obstacle instead of about its centre … you have to find what energy it needs to just reach "vertical". :wink:
 
Ok, how you say then I must calculate energy whith formula E=mgh, to determine what power I need to pass an obstacle with "h" height.

But, how is explained that a wheel with bigger radius pass a same obstacle easier than one with smaller?

How much role play differences in dimentions, is there something with balance.
 
vladim1 said:
But, how is explained that a wheel with bigger radius pass a same obstacle easier than one with smaller?

How much role play differences in dimentions, is there something with balance.

uh-uh, you tell me :wink:

for a wheel of density ρ, radius r, and speed v, what is its energy, and so what is the maximum height h of an obstacle that it will clear?

and how does that h depend, dimensionally, on ρ, r, and v, separately? :smile:
 
Thinking we discuss about thing I don't need.

Me interest is to calculate rolling resistance (in N) for a surface which have obstacles of 1cm height (eg. after each 5cm). Also I observe that a heighter wheel go easier after road's deformations.
 
I found a formula to calculate force needed to aply on axle (pushed wheel) to overcome obstacle:

F=m*g*sqrt(2rh-h^2)/(r-h)

where:

m= load
r= radius of wheel
h= obstacle height

Which formula must I use to calculate torque needed to overcome same obstacle, rotation start from rest wheel.

Can I use T=F*r ?

F - force calculated with above formula
r - wheel radius.
 
T = F*(r-h)

To derive the formula that you found, simply take the sum of the moments about the tip of the obstacle.

sum of Moments: F*(r-h) - mg*sqrt[r^2 - (r-h)^2] = 0
 
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