Created Date: September 13,2012

Created by: Tinker

I'm trying figure out how to size a motor for a vechicle drag chain/conveyor.

I need 40-60 FPM and enough power to carry 8 cars at one time.

The chain has a pitch of 2.61 inch and the sprocket has is a 12 tooth.

By my math that makes it a 10 inch Dia Sprocket.

So in order to get 50 FPM (Middle ground) I need 60 RPMs at the sprocket.

12 * 2.61 " = 31.32 " or about 2.5ft (or 10 " * pi = 31.4 ")

2.5ft/revolution * 60 rpm = 150fpm

50 fpm would be aout 20rpm (1800 / 20 = 90:1 ratio)

I want to use a 10 HP 1,800 RPM Motor so using a 20:1 gear box I get a needed motor to spin at 1,204 RPMs

Heres where I run in to trouble.... I need to know how much force I'm going to get at the chain.

I know that to find Tq from HP i need RPM. So lets say I get that 10 HP and I run it at 1800 RPM

(HP * 5252)/RPM

(10 * 5252)/1800 = ~30 ft-lbs

However, what happens to the HP rating if I drop it to 1200RPM? Do I keep the constant 10 HP or do I drop it say proportionally? 0 HP @ 0 RPMs, 10 HP @ 1800 RPMs then at 1200 RPM would I get 6.6HP?

If that works then might the formula would be?

(6.6 * 5252)/1200 = ~28 ft-lbs

If that works then I would get 584 ft-lbs a the sprocket shaft and 702 lbs of thrust from the chain.

Thanks for the help!

Over a resonable speed range you can get constant torque (diminishing horsepower) To save myself some math I just looked up the specifications for the 10hp Blackmax Y547 sold by AD and for it, the full load torque is 29.5 lb-ft (breakdown torque is 125lb-ft, something to keep in mind when selecting the gear box)

at a 20:1 rediction that would be nominaly 590 lb-ft, but I'm guessing you'll be using a worm gear and they are not super effcient for a 20;1 ratio 80% might be a resonable first appoximation 590 * 80% ~470 lb-ft

you sprockets 5 " radious = about 0.42ft

470/0.42 ~ 1100lb.

A higer ratio worm gear will be even less effcient, but a 60:1 at 70% would give you close to 3000lb

Created Date: September 13,2012

Created by: dieseltwitch

Thank you! Its a little bits that mess me up!

Ok so what do you think would be more efficient. a worm drive or a hydraulic motor-pump combo?

the other problem Im trying to solve is how much power do I actually need.

How can Tq remain constant while horsepower is diminishing? If HP is a function of Tq and rpm. if you reduce one doesn't the other one change along with it. Im shooting in the dark as to the power needed but I think I can do with a lot less than 10 hp. a 2HP motor would give me 1,280 ft-lbs (not counting for the eft losses) to me that would be enough to push 8 cars in neutral... yes no? how the heck do I calculate that?

Created Date: September 13,2012

Created by: Tinker

Thank you! Its a little bits that mess me up!

Ok so what do you think would be more efficient. a worm drive or a hydraulic motor-pump combo?

over all I suspectthe worm gear will still be more effcient than hydraulics, but there may be other factors to recomend hydralics. To handle a 125lb-ft breakdown torque at the input youd be looking at one beefy worm gear, not that the hydraulis motor would be all that small either.

the other problem Im trying to solve is how much power do I actually need.

How can Tq remain constant while horsepower is diminishing? If HP is a function of Tq and rpm. if you reduce one doesn't the other one change along with it. Im shooting in the dark as to the power needed but I think I can do with a lot less than 10 hp. a 2HP motor would give me 1,280 ft-lbs (not counting for the eft losses) to me that would be enough to push 8 cars in neutral... yes no? how the heck do I calculate that?

"If HP is a function of Tq and rpm " has three variables, one of them can remain constant, if torque is constant and speed changes then hp has to change. If Hp were to be constant then, yes, the torque would have to change with the speed, but if the hp is allowed to fall the torque can remain constant with speed change (over a 1000:1 ratio for the Blackmax TENV).

"Would be enough to push 8 cars in neutral... yes no? how the heck do I calculate that? " I have no idea, when math fails me I go empirical. I'd go to a sporting good store and buy a fish scale and a stop watch and start dragging cars around

P.S. make sure you try dragging a Hum-vee with underinflated tires to get near worst case (you may need to get some help to get up to steady 50fpm)

Created Date: September 13,2012

Created by: dieseltwitch

Ok well in order to get the 90:1 ration that would be a massive worm gear setup anyways. So I 've been looking into planetary. they advertises 96% Eff... Realistic?

Your description makes a lot of sense on HP & TQ. Got me thinking back that TQ is work, and HP is the speed that work is being done. so it makes sense that tq can stay up while hp falls with RPM's

Created Date: September 13,2012

Created by: dieseltwitch

So i found this great formula and Rolling Resistance Coefficient Chart

HERE

So taking a worst case scenario of 8 truck each weighting 8000 lbs (the curb weight of an F350 ) with the worst listed RRC (Rolling Resistance Coefficient) of .03 (car tires on tar or asphalt) I get a needed 240 lbs of force needed to move the trucks. for a total of 1920 lbs of force. that means I need just a 5 HP motor in a worst case.

However, I don't think that motor would really ever run at that much of a load. if the chain was loaded with 8x 3000 lbs cars with RRC of .01 i would only need 240 lbs of thrust from the chain, or 1 roughly a 1 HP load? Sound about right?

I love having people to bounce ideas off of

Created Date: September 13,2012

Created by: Adisharr

I haven't looked at any of the links but most of your force is going to be accelerating from a stop. Assuming the car wash works like others in my area, occassionally all the cars come to a stop and then start moving again.

You 're also grabbing the last vehicle and accelerating from a dead stop to the current chain speed. Just some factors to consider.

Created Date: September 13,2012

Created by: dieseltwitch

What about the force needed to push the brushes out of the way? Wouldn't the the counter rotating forces need to be accounted for?

I haven't looked at any of the links but most of your force is going to be accelerating from a stop. Assuming the car wash works like others in my area, occassionally all the cars come to a stop and then start moving again.

You 're also grabbing the last vehicle and accelerating from a dead stop to the current chain speed. Just some factors to consider.

Good points. Right now I have a car wash system up and running that I converted to VFD's. They have a 20HP motor that used run 8 sets brushes + the conveyor. Now it only runs the conveyor and 1 set of brushes. When a car is loaded the motor load fluctuates only about 2% and nothing is noticeable when it goes through the brushes.

I just did some more math. Check it for me?

If I have 8x 8000 lbs trucks on the line all at a stop (mass = ~30,000 kg) and I need to accelerate them to a speed of 83 fpm (200 cars per hours) (velocity = 0.42 m/s) with an available 2,840 lbs of thrust at the chain using a 5 HP motor (force = 12,660 N) then I can accelerate all the trucks to speed in ~1 (0.995) second?

I used the formula f = m*(v/t)

I reworked it to solve for time t = (m*v)/f

Does that work right?

Created Date: September 14,2012

Created by: Tinker

If I have 8x 8000 lbs trucks on the line all at a stop (mass = ~30,000 kg) and I need to accelerate them to a speed of 83 fpm (200 cars per hours) (velocity = 0.42 m/s) with an available 2,840 lbs of thrust at the chain using a 5 HP motor (force = 12,660 N) then I can accelerate all the trucks to speed in ~1 (0.995) second?

I used the formula f = m*(v/t)

I reworked it to solve for time t = (m*v)/f

Does that work right?

well, sort of, there is one little mistake.

Personally, while I’d prefer to use SI units when I can, if a project starts with "United States customary units " I prefer to stick with them rather than do a bunch of conversions (you might ask NASA about mixed units).

So, sticking with the feet and pounds (force) used in the original post:

2,840lbf * 83 fpm = 235,720 ft-lb/min or 3929 ft-lb/second

By definition one horsepower is 550 ft-lb/second

3929 / 550 = 7.14, allowing for inefficiency you are back to the 10hp motor of the original post, not 5hp. Of course since this is for a relatively brief period you can operate the motor at somewhat more than it’s continuous torque rating, but I’d hesitate to use 200% as the design criteria.

Given that a pound force is defined as the force required to accelerate a pound mass at (about) 32.2 ft/second^2

2,840lbf would accelerate a 64,000lb mass at about 1.43 ft/s^2 or 85.7 fpm/second so your ~ 1 second is correct for a 10hp motor, so about 2 seconds for a 5hp. But how fast do you even need to accelerate? A couple of seconds sounds reasonable to me.

However, those numbers only apply in the absence of friction.

So taking a worst case scenario of 8 truck each weighting 8000 lbs (the curb weight of an F350 ) with the worst listed RRC (Rolling Resistance Coefficient) of .03 (car tires on tar or asphalt) I get a needed 240 lbs of force needed to move the trucks. for a total of 1920 lbs of force. that means I need just a 5 HP motor in a worst case.

2840 – 1920 leaves less than a thousand for acceleration, for an acceleration time in the 3-4 second range, still sounds reasonable to me. And that is for your worst case loading (but a 10hp motor). Note that a 5hp motor runing at it's continous rating won't really be able to accelrate your worst case load at all, and while you can run the motor well above it's continous rating for a short time, the accelration time might not be so short with a 5hp motor so caution is advised. In another post you asked about gear drive vs. hydraulic, this is a case where a variable dispalcement hydraulic drive may have advantages (the cheaper constant dispalcement not so much), but the hardware could be expensive and overall effciency might not be so good (quite posibly more expensive and less effcient than using a 10hp motor and running it at less than half it's capacity most of the time)

Created Date: September 14,2012

Created by: dieseltwitch

well, sort of, there is one little mistake.

Personally, while I’d prefer to use SI units when I can, if a project starts with "United States customary units " I prefer to stick with them rather than do a bunch of conversions (you might ask NASA about mixed units).

So, sticking with the feet and pounds (force) used in the original post:

2,840lbf * 83 fpm = 235,720 ft-lb/min or 3929 ft-lb/second

By definition one horsepower is 550 ft-lb/second

3929 / 550 = 7.14, allowing for inefficiency you are back to the 10hp motor of the original post, not 5hp. Of course since this is for a relatively brief period you can operate the motor at somewhat more than it’s continuous torque rating, but I’d hesitate to use 200% as the design criteria.

Given that a pound force is defined as the force required to accelerate a pound mass at (about) 32.2 ft/second^2

2,840lbf would accelerate a 64,000lb mass at about 1.43 ft/s^2 or 85.7 fpm/second so your ~ 1 second is correct for a 10hp motor, so about 2 seconds for a 5hp. But how fast do you even need to accelerate? A couple of seconds sounds reasonable to me.

However, those numbers only apply in the absence of friction.

2840 – 1920 leaves less than a thousand for acceleration, for an acceleration time in the 3-4 second range, still sounds reasonable to me. And that is for your worst case loading (but a 10hp motor). Note that a 5hp motor runing at it's continous rating won't really be able to accelrate your worst case load at all, and while you can run the motor well above it's continous rating for a short time, the accelration time might not be so short with a 5hp motor so caution is advised. In another post you asked about gear drive vs. hydraulic, this is a case where a variable dispalcement hydraulic drive may have advantages (the cheaper constant dispalcement not so much), but the hardware could be expensive and overall effciency might not be so good (quite posibly more expensive and less effcient than using a 10hp motor and running it at less than half it's capacity most of the time)

Does your numbers take into account that power goes through the gear box? Im just trying to figure out how all the numbers work together when you put so many parts in the equation. Right now I'm sitting on 90:1 gear box... that has to account for something?

I have thought about the Hydro but while it works good when it goes bad it cause a lot of problems, leaks, messy......