Torque required to Move Wagon (4 wheel on a frame)

Assuming you're getting u from a 30° slope, looks OK. Suggest you add a generous amount of safety margin - 25 to 50% maybe (wouldn't want to get stuck with that much garbage to shovel out!).

May be a daft question, but can't you use a fixed winch & cable? Steerable trolley, maybe?

Another point - what kind of motor & control are you considering? That's a hell of a lot of torque for direct drive (i.e. no gearbox).

Thanks JohnDG for replying,

The 0.5 I used was for friction (Steel wheel on the Concrete) that number was a "WAG", the concrete surface is flat no incline or anything. I was thinking about using a winch the numbers I came up for the Force required for the winch to pull is (Force)(Friction) = 1700*.5 = 850 lbs thats the force required to pull the 1700lbs wagon on 4 wheel, But I don't have a specific spot to attach the fix winch since I move the bin around, but your right If my calculations are correct it would be cheaper to use a fix winch, only thing it pulls in one direction.

Samantha

Your 0.5 as a 'finger in the wind' sounds OK.

Another problem you'll have to consider is getting power to the trolley.

Hi Samantha,

If you still prefer to use a fixed winch, you can pull the wagon in the opposition direction by running the cable over a pulley to the reversed starting point. I presume you would need this to return the empty trolley to the starting point.

How far are you looking to move the trolley overall? Would a boat trailer winch running from a car/truck battery supply be adequate?

If all you needed to do was move it back and forth, you could set up something like a garage door opener. This could still work if you have to move the bin around, as long as the bin moves in a consistent pattern. You could even use an old door opener by upgrading the motor, that way you have a remote control.

Actually a big door opener might be just what you need even if you cannot use it like the pulley and chain design. Mount the door opener and use a bicycle gear setup to vary the speed/torque settings. Plus, you get the wireless controller! If you can't get it to run on mains (household) electricity, swap the motor out for an efficient battery convenient motor and use the bicycle gears to adjust the ratio.

Drew

Since this wagon is to move horizontally, you do not need to lift this load--only accelerate it to the modest speed given. The friction you need to overcome is only that of the wheel/motor bearings.

An acceleration of g/20 should be adequate. In effect, this means dividing the wagon+load weight by 20, and then multiplying by the wheel radius to get required torque ≈ 1700/20 x 3 = 255 lb-in. (If you want faster acceleration than that, adjust the computation accordingly, and perhaps multiply by a safety factor.)

However, if you don't want to "peel out," you need to check that the sideways force exerted by the wheels (85 lb) won't make them slide. That would take 1700 lb x 0.5 coeff. fric. = 850 lb, which is way more than the contemplated motors will do; so you're safe there.

Interesting project, btw. What kind of motors and power provision are you thinking of?

An electric pallet jack motor might be an excellent candidate for this application.

I accepted the figure 0.5 on the basis that the floor may be lumpy & maybe have the odd bit of garbage that the trolley would need to 'climb over'. I agree your figure should be plenty in ideal circumstances.

Good point. Even a little bump of 30° would require lifting 1/4 to 1/2 of the load over the bump. (Discounting inertia even if the wagon is at speed.)

The 0.5 figure pertains both to estimated coeff. fric. and to climbing a 30° slope. Those are separate issues.

Hey guys,

Thanks for the reply, for power I was considering using a hydraulic motor but am now considering using an electric pallet jack motor as per Tornado reply, thanks Tornado and JohnDg.

I was having this discussion with my friend and we seem to disagree on how I came up with the torque calculation, I will explain the discussion and let me know what you all think.

I am using (2) motors to drive the wagon and they are attached to the "Two Back Wheel", if the load is equally distributed each wheel will see a load of "425 lbs" (1700/4). I calculated the torque using: [(Load per wheel)(Friction)(Wheel Radius)] x (2)] this will be the torque required per motor (1280 lb-in), I multiply by (2) because I took the torque of the front wheel and added it to the torque of the rear wheel.

The disagreement is my friend thinks that I should NOT multiply by (2) because the rear wheel is only seeing 425lbs and the front wheel has no input on the torque of the rear wheel, therefore the torque required for each motor is only (640 lb-in), I somewhat disagreed because even though the (2) back wheel is driving the wagon it still has to move 1700lbs load. What do you all think, am I right or wrong?

Tornado can you let me know the formula you used to acquire the 255lb-in, I notice you used Torque=(Force/a)*(wheel radius). I am trying to figure out that formula but the closest I came up with is Force=(m)(a), I am just trying to justify the 255lb-in because that number is much lower that what I go and works best for me, I just need to understand it a little bit better.

Thanks guys,

Samantha

This is because, except for going over presumably small bumps, you are not lifting the 1700-lb load; you are only accelerating it horizontally.

For an analogy, think of putting a 3000-lb car in neutral and pushing it. The main things you are overcoming are tire flexure and wheel bearing friction. About 100-200 lbf is enough. Even locomotives can be pulled by a single strong person, but there is virtually no wheel flexure, and only bearing friction (probably roller bearings) is involved.

Accelerating 1700 lb at g/20 is the same as accelerating 85 lb at 1 g, which takes 85 lb., applied at 3" radius, hence 255 lb-in of torque. You might "bump" this up some to help in getting over small bumps on the travel surface. A pallet jack motor is already sized almost perfectly for this task. A common power source is 24 VDC batteries; this and the motor will add a bit of weight to the wagon assembly.

Thanks Tornado, That information was very usefull, what do you think about the discussion I posted ealier, do you think I was correct in my thinking?

Samantha

I like wheelchair motors. Not quite as strong as a pallet motor, but more common and probably easier to find used.

Drew

Good idea. A golf cart motor might also work. All of these options seem to be about in the right torque/speed range. It is a matter of availability and cost.

If any of these motors come attached to a single-wheel drive, the cart could be reconfigured as a 3-wheeler. (And even steerable with a linear actuator matching the motor voltage.)

What do you guys think of the discussion I posted earlier?

"I was having this discussion with my friend and we seem to disagree below is the discussion:

I am using (2) motors to drive the wagon and they are attached to the "Two Back Wheel", if the load is equally distributed each wheel will see a load of "425 lbs" (1700/4). I calculated the torque using: [(Load per wheel)(Friction)(Wheel Radius)] x (2)] this will be the torque required per motor (1280 lb-in), I multiply by (2) because I took the torque of the front wheel and added it to the torque of the rear wheel.

The disagreement is my friend thinks that I should NOT multiply by (2) because the rear wheel is only seeing 425lbs and the front wheel has no input on the torque of the rear wheel, therefore the torque required for each motor is only (640 lb-in), I somewhat disagreed because even though the (2) back wheel is driving the wagon it still has to move 1700lbs load. What do you all think, am I right or wrong?"

I just want to know if I am correct in my assumption, but oreng78 has a good point, if I use (2) wheel there is a possibility to go in circles, so I am going to use (1) shaft to drive both rear wheels.

Samantha.

Hello Blink,

Your reply is absolutely prefect, thanks for the info. I have considered using v-groove wheel and using angle iron as my tracks so the wagon will be moving on the angle iron similar to this: (http://www.castercity.com/vgroovew.htm#Table%20-%20Casters"

I taught of using the angle iron as tracks because just in case junk falls on the track at least it will fall of the angle iron peak an not get caught on the wheel. I am also considering using a winch drive I am drawing a sketch and will be comparing the prices (motor vs winch) and will probably choose the cheaper method. Thanks for your post it is very helpful.

Thanks,

Samantha

Hello Blink,

I was just going through your numbers and I got different values. For example:

For Tractive effort I got (1700 lbs/32 = 53.125 lbs not 218 lbs) and a 30% increase of 53.125 is 70 lbs. Am I doing something wrong with the numbers?

Samantha

For Tractive effort I got (1700 lbs/32 = 53.125 lbs not 218 lbs) and a 30% increase of 53.125 is 70 lbs. Am I doing something wrong with the numbers?

Actually, we're both doing something wrong. 1700/32 is indeed 53.125. This is the tractive effort required to accelerate from 0 to 1fps in one second. ( I can't imagine where I got 218.)

But the 30% is not an increase, it is a grade: 3 feet rise in 10 feet of run. The "grade is really a micro grade encountered in climbing over little pieces of debris.

The tractive effort required to climb a grade is equal to the grade percentage times the vehicle weight, for grades that are not very steep. (It's handy for estimating.) In fact, 30% (16.7 degrees) is about as steep as you would want to go without using the more precise (sin of the grade angle x vehicle weight). In this case, the sin of 16.7 degrees is .287, close enough to .3 for estimating purposes. Sticking with .3, then the tractive effort required to climb the 30% grade (over some piece of crud) would be .3 times the vehicle weight: 510 lb, if all the wheels are simultaneously climbing over stuff.

The size of the piece of crud to climb over and the size of the wheel affects the apparent grade. The wheel axle will follow an arc, and the tangent to that arc is the apparent grade at the instant that the obstacle is encountered. The smaller the obstacle relative to the wheel size, the smaller this tangent angle. If your frame were perfectly stiff, then the weight lifted by one wheel going over an obstacle would be half the vehicle weight. It is very unlikely that the frame would be this stiff, for small obstacles, so the actual weight lifted at one wheel would be not too much more than 1/4 the weight of the whole cart. How many wheels might encounter an obstacle at once? If you have a lot of loose gravel that finds it way into the pit, then perhaps half the wheels might be obstructed at the same time. On the other hand, if the wheels are on tracks, then there may never be any obstacles of concern, in which case, the tractive effort might not have to be more than that required for acceleration plus 10 lb for rolling resistance, plus a safety margin.

hey samantha, i see that here u are equating the torque as on the point of contact on ground hence u r considering "u". well u may have learnt that friction is self adjusting and will match the value of force at that point of contact and if ur torque creates a greater force on that point( u*f*r > u*m*g) then the wheels will slip/slide and ur vehicle will not move. my question to u is wether ur vehicle moved or not with ur calculations, as this reply is after 4-5 months... , thnx titan

Yes it did move, actually it worked out quite well.