Optimum Wheel Radius for a Terrestrial Robot

[Christian Hoermann]

Hello everyone! This is my first post here so please excuse me if I don't have the format right yet.

Background: I'm a Mechanical Engineering student working on a robotics team and I'm tasked with designing the wheels. The robot is currently using 5 in radius wheels with old motors.

The givens: We already have specced new motors to work with. We are limiting the specs to be maxed out at the following:
Torque: 806.4 oz-in Angular Velocity: 140 RPM Current: 20.16 A
The weight of the robot is 110 lbs (1760 oz) distributed over 4 wheels.

Problem: I have been tasked with calculating the optimal wheel radius to go up amaximum of a 45 degree incline at a rate of 4.5 mph (yes I know it's specific) neglecting friction for now.

My attempts:
1) Assuming constant velocity (no acceleration), the free body diagram indicates that each wheel will experience a resolved force of Fr=(m/4)g*sin(theta) where theta is the incline of the slope.
I am using 386.09 in/s^2 for g. This yields Fr=120,123 oz-in/s^2 which is already setting off red flags
since the equation for torque is t=F*r,
I calculated that the minimum wheel radius is 806.4/120123 = 0.0067 in.
I obviously am very hesitant to tell my lead that the radius can be reduced to 7 thou.

2) The arc length formula for a nonslipping, rolling body yields that Velocity=radius*angular velovity (v=r*w)
so I tried to calculate it using this formula.
I used r=V/w = (4.5mph/140rpm)=(79.2in/s / 14.66rad/s) = 5.402 in. This looks much better, but does not take into account the wieght of the robot or the angle of the slope so it has me very concerned.

3) I was very tired when doing this one; so bear with me please.
Power=Force*Velocity= (m/4)g*sin(theta) * v
Also, Power = Torque * Angular Velocity
Therefore Force*Velocity=Torque*Angular Velocity
Leading to Angular Velocity = Force*Velocity/Torque
And v=rw
so we're back to r=T/F which is what the first equation got me.

Again, this is my first post here so I am very sorry if this is in an improper place/format.

 

[phinds]

Just so you know, your subject line says you are looking for a minimum wheel radius but your post says you have been tasked with finding an optimum wheel radius. You might want to ask a moderator to change your subject line so there's no confusion. (we members can edit our posts for a limited time after posting but we can't edit the subject line)

 

[Dr.D]

Don't forget that the wheel must support the machine above, and thus must not penetrate the surface it is running on. Just because a pin head might be adequate for speed, that would not make it a good choice for support. What sort of surface will this operate on? Dirt, gravel, concrete, etc?

 

[JBA]

Forgive me if the information is some where in your calculations; but, I have simply been unable to ferret it out. How many driving motors does your robot have? Also, are your motor(s) direct drive at the wheels(s); or, do have any gear reductions in your drive train?

 

[Christian Hoermann]

Hey guys, thanks for getting back so quickly! I'll try to answer your questions in order.

phinds: thanks for the advice! I'll certainly see about that.

Dr. D: This is an all-terain robot meant to simulate Mars Rovers in a national competition. We will be operating on all sorts of terrain. However, we will also be expected to drive it around our university to promote our department. Therefore it has to function on smooth, hard surfaces. Also, the wheel (once manufactured) will be coated with our 11 herbs and spices polymer coating. That's why I am disregarding friction for now.

JBA: That's ok, I guess I didn't elaborate. The robot will be powered by 4 gearmotors and the input/output of the whole assembly is what I submitted above. If you're interested, its this gearmotor: http://www.andymark.com/product-p/am-pgseries.htm the PG71 gearbox with the 9015 motor. Unfortunately, Andymark doesn't provide the full speed-torque curve. So I extrapolated the data given for stall load and for no load. I understand that this assumption assumes linearity, but it's the best I can do without having buying them and performing experiments.

 

[berkeman]

Just so you know, your subject line says you are looking for a minimum wheel radius but your post says you have been tasked with finding an optimum wheel radius. You might want to ask a moderator to change your subject line so there's no confusion.

Thread title edited.

 

[JBA]

You should realize that since your vehicle weight, speed and slope, and available torque are already determined, a trial and error loop solving process (or possibly simultaneous equations) will be required for the solution to your problem because both the drive torque required and the speed of the robot are directly related to the radius of the driving wheels. As the wheel radius increases both the speed and the driving torque required simultaneously increase.

 

[Christian Hoermann]

Thanks berkeman, I was actually looking for staff when you posted that.

JBA: I understand that the wheel radius affects the robot's linear speed and the resulting linear force that the wheel outputs onto the ground. The torque, however, is limited by the max allowable current (which we are limiting to the originally posted values). I am looking for a way to derive the critical radius size that will allow us to meet the requirements.

 

[CWatters]

On rough terrain you generally want large wheels.

 

[Nidum]

On rough terrain you generally want large wheels.

+1

One measure of an all terrain vehicles capabilities is the maximum height of a square edge step that it can climb over or descend from .

Given a specified maximum step size the minimum wheel diameter required can be decided easily .

For realistic step heights this generally means large diameter wheels .

The drive power required for a climb over and braking power required for a descent can then decided by separate calculations .

 

[jack action]

1) Assuming constant velocity (no acceleration), the free body diagram indicates that each wheel will experience a resolved force of Fr=(m/4)g*sin(theta) where theta is the incline of the slope.
I am using 386.09 in/s^2 for g. This yields Fr=120,123 oz-in/s^2 which is already setting off red flags
since the equation for torque is t=F*r,
I calculated that the minimum wheel radius is 806.4/120123 = 0.0067 in.
I obviously am very hesitant to tell my lead that the radius can be reduced to 7 thou.

This is wrong (thus the red flags).

Your weight is $W = 1760\ oz_f$ (ounce-force), which is a force, where $W = mg$. Thus:
$$F_r = \frac{mg}{4}\sin\theta$$
$$F_r = \frac{W}{4}\sin\theta$$
$$F_r = \frac{1760\ oz_f}{4}\sin45$$
$$F_r = 311\ oz_f$$
And:
$$r = \frac{T}{F_r} = \frac{806.4\ oz_f.in}{311\ oz_f} = 2.59\ in$$
Furthermore, the capability to climb a slope is also determined by the vehicle characteristics, especially the wheel friction coefficient (regardless of the torque input). More info here.

 

[Mech_Engineer]

For a start, the maximum power output for the motor gives you a bound for the fastest you can drive up the incline. The link you give specifies a power output of 100 W per gearmotor, so 400 W total. The equation for calculating max speed up an incline is:

$$v_{max}=\frac{Power}{Mass*g*sin(angle)}$$

That calculates out to about about 1.27 m/s, or about 2.8 mi/hr. So, your current setup doesn't have enough power to drive up at 45 degree slope at the specified speed...

 

[Mech_Engineer]

We already have specced new motors to work with. We are limiting the specs to be maxed out at the following:
Torque: 806.4 oz-in

Also, the link you provided specs the stall torque of your gearmotor at 14 ft*lbf (2688 in*ozf). Do you have any clarification on the discrepancy?

 

[JBA]

@Mech_Engineer, I have difficulty with metric, but it appears to me that you may be off by one decimal point because I get 12.61 m/s

 

[Mech_Engineer]

@Mech_Engineer, I have difficulty with metric, but it appears to me that you may be off by one decimal point because I get 12.61 m/s

Thanks for checking my numbers, the numbers I'm using are [[updated for 110 lbm rover]]:

Power: 400 W (4x 100 W motors)
Mass: 49.89 kg (110 lbm)
g: 9.81 m/s^2
Angle: 45 deg

Calculates out to:
$$v_{max}=\frac{400}{49.89*9.81*sin(45)}=1.16\frac{m}{s}=2.59\frac{mi}{hr}$$

 

[JBA]

Note: The real US weight given is 110 lbs and 110 lb weight = 49.9 kg force not kg mass, as I understand it. But apart from the different weight value. In order to get the mass of 110 lbs in US units requires dividing the weight by 32.18 ft/sec^2 = 3.42 slugs if you want to accelerate it by a different acceleration value.

 

[JBA]

@https://www.physicsforums.com/members/christian-hoermann.629984/, Attached is a copy of my excel solution at this point for your review.

 

[jack action]

Note: The real US weight given is 110 lbs and 110 lb weight = 49.9 kg force not kg mass, as I understand it. But apart from the different weight value. In order to get the mass of 110 lbs in US units requires dividing the weight by 32.18 ft/sec^2 = 3.42 slugs if you want to accelerate it by a different acceleration value.

$1\ kg_f = 1\ kg \times g$, just like $1\ lb_f = 1\ lb_m \times g$.

So $49.9\ kg_f = 49.9\ kg \times g$. But $1\ kg_f = 9.0665\ N$ - which is the basic force unit to be used with $mg$ - thus $49.9\ kg_f = 489.35\ N$.

With the US system, $110\ lb_f = 110\ lb_m \times g$. But $1\ slug = 1 \ lb_f.s^2/ft = g\ lb_m.s^2/ft = 32.174\ lb_m$ - which is the basic mass unit to be used with $m$ - thus $110\ lb_m = 3.42\ slug$.

 

[Christian Hoermann]

This is wrong (thus the red flags).

Your weight is $W = 1760\ oz_f$ (ounce-force), which is a force, where $W = mg$. Thus:
$$F_r = \frac{mg}{4}\sin\theta$$
$$F_r = \frac{W}{4}\sin\theta$$
$$F_r = \frac{1760\ oz_f}{4}\sin45$$
$$F_r = 311\ oz_f$$
And:
$$r = \frac{T}{F_r} = \frac{806.4\ oz_f.in}{311\ oz_f} = 2.59\ in$$
Furthermore, the capability to climb a slope is also determined by the vehicle characteristics, especially the wheel friction coefficient (regardless of the torque input). More info here.

Ah, I can't believe it was such a simple mistake. Thank you!
The wheel design is actually a cylinder with an aluminum frame and a polymer coating (think along the lines of NASA's Curiosity rover, but without the damage).
We will change various factors to improve friction and sliding until we can reach the best grip while still being a fairly easy ride on smooth surfaces.

For a start, the maximum power output for the motor gives you a bound for the fastest you can drive up the incline. The link you give specifies a power output of 100 W per gearmotor, so 400 W total. The equation for calculating max speed up an incline is:

$$v_{max}=\frac{Power}{Mass*g*sin(angle)}$$

That calculates out to about about 1.27 m/s, or about 2.8 mi/hr. So, your current setup doesn't have enough power to drive up at 45 degree slope at the specified speed...

Thank you very much! I will pass it along the chain. More importantly though, I understand how you calculated it so I can use that design formula in the future. I advised my superior to use another motor that provides more torque, but it was too slow for us even of flat ground.

Also, the link you provided specs the stall torque of your gearmotor at 14 ft*lbf (2688 in*ozf). Do you have any clarification on the discrepancy?

The link I provided is the distributor and (unfortunately) does not include the complete speed-torque curve. As I mentioned before, I took the values for rpm, torque, and current and devised my own. However, this is not very accurate because it assumes linearity (which is a pretty bad assumption from what I've read). I instructed excel to create the data along the increments of 10 rpm. That is why the torque and current aren't pretty numbers like the rpm. We knew that our internals cannot handle more than 20 A of current (wire thickness and fuses), so we chose the closest amount of current to that on the chart (20.16A) and got the rpm and torque from there.

 

[Christian Hoermann]

@https://www.physicsforums.com/members/christian-hoermann.629984/, Attached is a copy of my excel solution at this point for your review.

That is some good work! Can you please explain the formula for "maximum wheel radius" ? I'm reading that it's the max torque output/required vertical force*12. I assume the 12 is a conversion factor, but I can't find where it's needed.

Edit: ah...well I guess I need glasses. You put the radius in inches where the torque is in lbs-ft. Nevermind.

 

[JBA]

First, after posting and on review I realized I actually have some errors in this calculation that need to be corrected with regard to the last HP calculation and torque calculation.

What is shown as Reqd Torque should be "Available Torque" regardless of wheel size and should be for 400 watts not 100 (which is per wheel). Then by ratioing that to the rewired torque for the selected wheel size I should be able to calculate the actual required HP/watts required at the 140 rpm speed.

With regard to the first wheel size that is simply a calculation to determine the maximum radius boundary based upon the specified maximum locked rotor torque and then that allowed me to calculate the speed at that wheel radius and subsequently ratio the speed at that radius to your required speed to get the new wheel radius for your speed.

Sorry for the errors. Once I have made my required error corrections I will post the corrected version for review

 

[JBA]

At this point I have updated and revised my initial excel calculation with few additions including the simple "Power required based upon load and speed" calculation that should have done immediately in the first place. It also includes a table of torque and rpm vs wheel radius for a selected range of wheel radii, none of which change the basic required drive power.

At the same time, as has been discussed, this is only the basic theoretical power and does not take into account the additional power that will be required due to drive system friction, etc.

The increase of torque of the drive at the lower rpm or that the increase in required torque with increased wheel radius will always exceed your currently selected motor/gear drive's capabilities. At the same time, your currently selected 100 watt motor/gear drive is way below your specified 12 v & 20.4 amp max limit.

Note: As to your question regarding the x 12 conversion of the wheel radius from ft to inches. You are correct that this an unnecessary step from a calculation standpoint. It is simply due to the fact the under the Imperial/US measurement standard any dimensions of length less than a foot are always stated in inches, never in decimal fractions of a foot.

 

[Mech_Engineer]

As I mentioned before, I took the values for rpm, torque, and current and devised my own. However, this is not very accurate because it assumes linearity (which is a pretty bad assumption from what I've read).

From what I understand, this is a valid method for modeling a DC motor's torque vs. speed curve. See here: MIT- Understanding DC Motor Characteristics (Section 3.1, Torque/Speed Curves)

From MIT website:

I instructed excel to create the data along the increments of 10 rpm. That is why the torque and current aren't pretty numbers like the rpm. We knew that our internals cannot handle more than 20 A of current (wire thickness and fuses), so we chose the closest amount of current to that on the chart (20.16A) and got the rpm and torque from there.

The important thing to keep in mind is that because the motor's torque vs. speed curve is linear, and motor's power output is a product of the motor's torque and speed (a.k.a. the area under the torque vs. speed graph), the Power vs. Speed curve is an upside-down parabola with a maximum in the middle. The most important takeaway of this is the motor's maximum power output does not occur at the same point as max torque.

See here: Section 3.2: POWER/TORQUE and POWER/SPEED CURVES

When I do the math and calculate power as a function of speed using the AndyMark motor's specifications for stall torque and free speed, the math works out. When you create a linear torque-vs-speed equation with appropriate max torque and free speed, the maximum power for this DC motor is 100 W at 100 RPM (half the free speed), and the motor's torque at peak power output is 9.491 N-m (1344 in-ozf). Current draw at 100W should be around 8.3 amps if its a 12V motor.

Building Torque vs. Speed equation:

Torque vs. Speed plot:

Power vs. speed equation:

Power vs. Speed plot:

 

[Mech_Engineer]

For more information, you should also take a look here: Robot Shop: Drive motor sizing tutorial.

This website also provides a robot drive motor sizing tool here: Robot Shop: Drive Motor Sizing Tool. The calculator appears to provide numbers that are in-line with the estimates I've made so far, for what it's worth. Note that the drive wheel radius needs to be large based on your available power output and rpm at peak power out (I put in 24 deg. incline since that's the max incline you can ascend at 2.0 m/s with 400 W of peak power output) [Edit: This also assumes 100% efficiency in your drive system, which I would recommend using something lower like 65% for some safety factor]:

Max incline based on power output:

Robot wheel size based on peak power torque from my calculations: