Max wheel size according to torque

  • Thread starter Thread starter Jones1987
  • Start date Start date
  • Tags Tags
    Max Torque Wheel
AI Thread Summary
The discussion focuses on determining the optimal wheel size for a robot to climb inclines, specifically addressing the relationship between torque, weight, and incline angle. It emphasizes using the total torque from all motors to calculate the maximum wheel radius, as this ensures adequate force to overcome gravitational pull on inclines. The conversation also highlights the importance of considering traction and the robot's flip-over limit, which occurs when the normal force on the front axle is zero. Participants clarify that calculations for wheel size on flat surfaces yield different results than on inclines, with the maximum radius theoretically being infinite on flat terrain. Understanding these dynamics is crucial for effective robot design and performance on varying slopes.
Jones1987
Messages
74
Reaction score
0
Hi, I'm researching into what will be the best size wheels to put onto a motor to allow my robot to climb an incline, and then possibly find out what is the maximum incline it can reach.

I'm using this thread as a source for the math:

http://forums.trossenrobotics.com/archive/index.php/t-2900.html

But where he states this

"So, given 47 oz-in of torque with four motors (1.3 Newton-meters), mass of 2 Kg and an angle of 15 degrees, the answer is:

r = 1.3 / (9.8 * 2 * 0.26) = 0.26 m"

Would you not use the torque of a single motor to find out its max wheel size? Rather than combine the total torque and use that?

Also, once I've found my ideal size of wheel to tackle a 15degree incline, I'm taking an educated guess that a wheel to combat a 20degree incline be a smaller?
 
Engineering news on Phys.org
If you assume that your robot has 4 wheels of radius r, each connected to a motor with torque T, the force balance equation is:

m*g*sin(theta) = (T/r)1 + (T/r)2 + (T/r)3 + (T/r)4 = 4*(T/r) = (4*T)/r

and 4*T is the sum of the torque from all motors. This is why you need the torque from all motors.

Don't forget that this is the maximum wheel radius for your robot to be able to climb that hill (actually it will be only just enough for preventing your robot to go downhill); you can (and should) go smaller.

They are also 2 other characteristics that will limit the climbing ability of your robot: its traction and its flip over limit.

Read this http://hpwizard.com/car-performance.html" for more info (look at the bottom of the page: Theory»Longitudinal acceleration»Accelerating»Hill climbing)
 
Last edited by a moderator:
jack action said:
If you assume that your robot has 4 wheels of radius r, each connected to a motor with torque T, the force balance equation is:

m*g*sin(theta) = (T/r)1 + (T/r)2 + (T/r)3 + (T/r)4 = 4*(T/r) = (4*T)/r

and 4*T is the sum of the torque from all motors. This is why you need the torque from all motors.

Don't forget that this is the maximum wheel radius for your robot to be able to climb that hill (actually it will be only just enough for preventing your robot to go downhill); you can (and should) go smaller.

They are also 2 other characteristics that will limit the climbing ability of your robot: its traction and its flip over limit.

Read this http://hpwizard.com/car-performance.html" for more info (look at the bottom of the page: Theory»Longitudinal acceleration»Accelerating»Hill climbing)

Hi jack, thanks for the detailed reply. I assume to obtain max wheel size, I just use sum of torque, divided by the force. Which will give me a max radius to use

So e.g.
m = 2kg
T = 0.355 * 2 (2 motors) = 0.71Nm
F = (9.8 * 2 * sin(15)) = 5.07N

So Rmax = 0.71 / 5.07
Rmax = 0.14m

Is this correct?

I'm looking at the flip over limit, and I don't know if it's because I've not long been awake, but I'm not grasping it. Will the resultant value of tan(theta) be the max angle before this robot will begin to lift and flip over?
 
Last edited by a moderator:
Yes, your calculations are correct.

As for the flip over limit, it is the angle at which the normal force on the front axle will be zero and the total weight of the vehicle will be on the rear axle. This is an unstable condition. At this point, the front axle can lift to any height and the vehicle may flip over.
 
jack action said:
Yes, your calculations are correct.

As for the flip over limit, it is the angle at which the normal force on the front axle will be zero and the total weight of the vehicle will be on the rear axle. This is an unstable condition. At this point, the front axle can lift to any height and the vehicle may flip over.

Ok brilliant. However for some reason, when I calc the ideal wheel size on a flat surface, I end up getting a lower number than what it would be on an incline. But surely on a flat surface you are allowed larger wheels?

R = T / (g * m)
R = 0.71 / (9.8 * 2)
R = 0.036m

So I have a feeling I'm missing something obvious here.
 
sin(0) equals 0, not 1. The value of Rmax is infinity on a flat surface.
 
Posted June 2024 - 15 years after starting this class. I have learned a whole lot. To get to the short course on making your stock car, late model, hobby stock E-mod handle, look at the index below. Read all posts on Roll Center, Jacking effect and Why does car drive straight to the wall when I gas it? Also read You really have two race cars. This will cover 90% of problems you have. Simply put, the car pushes going in and is loose coming out. You do not have enuff downforce on the right...
I'm trying to decide what size and type of galvanized steel I need for 2 cantilever extensions. The cantilever is 5 ft. The space between the two cantilever arms is a 17 ft Gap the center 7 ft of the 17 ft Gap we'll need to Bear approximately 17,000 lb spread evenly from the front of the cantilever to the back of the cantilever over 5 ft. I will put support beams across these cantilever arms to support the load evenly
Thread 'Physics of Stretch: What pressure does a band apply on a cylinder?'
Scenario 1 (figure 1) A continuous loop of elastic material is stretched around two metal bars. The top bar is attached to a load cell that reads force. The lower bar can be moved downwards to stretch the elastic material. The lower bar is moved downwards until the two bars are 1190mm apart, stretching the elastic material. The bars are 5mm thick, so the total internal loop length is 1200mm (1190mm + 5mm + 5mm). At this level of stretch, the load cell reads 45N tensile force. Key numbers...

Similar threads

Replies
4
Views
2K
Replies
23
Views
4K
Replies
8
Views
2K
Replies
1
Views
3K
Replies
3
Views
2K
Replies
6
Views
3K
Back
Top