Calculating Your Boat Winch's Pulling Capacity for Non-Mathematicians

[hownddog]

I am trying to determine what weight my boat ramp drum winch can pull. it has an electric 1HP motor that drives the drum via a chain and sprocket through a gearbox. what info is required to work out its capability. sorry as I am no mathematician but I do know there is a lot of variables involved and I can supply a lot of information that might be asked. appreciate if someone can help

 

[BvU]

Hi houwnd,

From googling Hp to kw you find 1 Hp $\approx$ 0.75 kW. So if it went without losses your motor could deliver 750 Joules per second.

Lifting 1 kg by 1 m/s costs 9.8 Joules per second (work against gravity). Round off to 10 J/s = 10 W.
So $m \times g \times v$ = 1 kg * 10 m/s² * 1 m/s = 10 kg m² s² = 10 J/s = 10 W.

You don't have to be a mathematician to manipulate this: for example, lifting 7500 kg by 1 cm/s comes out at 7500 kg * 10 m/s² * 0.01 m/s = 750 W.
That's the basics.

The ramp helps: you don't have to lift vertically but up a grade. What comes in is the sine of the angle with the horizontal: a 30 degree slope (sine(30)=0.5) makes it twice as easy (but of course your 1 cm/s along the slope translates to 0.5 cm/s vertically -- a bit slow, perhaps ?)

And friction doesn't help: you lose in the motor (but perhaps the 1Hp is power delivered on the axle, in which case that loss is accounted for), in the gearbox (probably considerably), in the boat trailer, possible pulleys, etc. And I haven't even brought up that to start the motion you need quite a bit of power to accelerate the load plus power to overcome static friction (usually more than friction while the stuff moves).

Does this help you on your way a little bit ?

 

[JBA]

In order to determine the maximum weight that can be started and pulled, it is necessary to know the "locked motor torque" of your motor (that must be obtained from the motor manufacturer), the motor's rated rpm, the sprocket diameter ratio of your chain drive and the gear ratio of your gearbox.

Edit: the diameter of your cable drum is also required.

All of this said, it sounds a s though you have a manufactured assembly and the manufacturer of that unit should be able to provide you with the units rated load.

 

[tygerdawg]

Look on the motor nameplate & find the output RPM.
Google the relationship equation for RPM, horsepower, & torque.
This is the output torque of the motor to the driving sprocket.
Via sprocket size ratios N1/N2 = S1/S2, you can get the torque on axle of the driven sprocket. Which is the torque applied to the take up spool. Call it "T".
Determine the max take-up diameter of the cable on the spool. Call it "r".
Max pulling force F = T / r. Which will equal your max load. Apply safety factors, friction loads, etc., as needed.

(somebody check my logic...awfully tired right now)

 

[hownddog]

Thanks for all replies. " BvU" let me know I have a capable motor so here is the specs on it ( 240 v,1HP, RPM1360,750watt ,4.35Amp) as to "tygerdraw's" info I think its ,torque 3.86 ft-lb/. gear ratio 3.45. max force 1.74 (not sure) . "JBA" has given me what info is needed apart from not being able to identify the manufacturer, so here's what I can supply:
Gearbox : 45-1
Gearbox drive sprocket: 11 teeth/ 60mm diameter
Drum sprocket: 38 teeth/200mm diameter
Drum diameter: 140mm
Drum length: 280mm( = 45 single wraps of 6mm cable)
Ramp gradient 10 Degrees
I understand, a change in gradient, static or rolling weight (witch it is) double wraps and adding pulleys etc. can all affect outcome

 

[JBA]

By combining the above information you can calculate the theoretical maximum operating load for your system; however, this calculated theoretical value may very well far exceed your system's actual maximum safe operating load because it does not take into account the maximum load rating of the gearbox or chain drive, system component's design load, the system mounting connections' load limit and the cable mfr's rated safe load, anyone of which can substantially limit the actual safe load for your system; plus an appropriate safety factor for the whole system in general.

 

[hownddog]

thanks for all the help everyone, but due to my limited schooling , I probably seem pretty dumb to you all sorry. so am I right in my previous post when I worked out max force or load = 1.75 (would this be tons or kilos ??) I agree with what JBA said as far as all the safety factors which I will take into account

 

[JBA]

Based upon my calculation the value @ 1 HP / 1360 rpm (rated hp and rpm) would = 1.19 tonnes (metric tonnes)

But, here is the catch, as the electric motor speed reduces its available torque increases (which is shown on the mfrs performance curve sheet for the motor), so the amount of motor torque and resulting pulling force can substantially exceed the 1.19 tonne force at the beginning of a pull, resulting in much higher gearbox and chain drive torques and pulling cable load.

This is the reason that I am very concerned about your using this type of calculation for judging your total pulling system's maximum safe operating load. What you are trying to do is basically design your system from scratch without having all of the necessary data sheets for all of the components of the system nor the background to calculate the strength limits of the units frame and fasteners design.

For this reason, I strongly recommend you make a serious effort to identify the manufacturer of your pulling system and get their design safe load rating for the unit. If you had simply stated that you had an unit rated for a given load then giving you the effect of a 10° ramp would be straightforward; but, what you are trying to do now far exceeds that.

 

[hownddog]

Thanks JBA, I realize your safety concerns but I am unable to identify the manufacturer. I know it is a professionally made unit but the info on it is illegible. it sits in a very solid 8mm galvanised welded plate steel cradle with 6 mounting bolt holes and weighs about 15 Kilograms. I know the cable is rated at 2585 kilo minimum breaking strength(which can be upgraded anyway) and the cradle will be chemical set bolted and fixed into solid 200mm thick concrete which will hold well over the weight. To put things into perspective I am going to use it to pull a combined 2000 kilo boat and trailer up a 10 degree ramp .From what I can gather with all the helpful info it can do this, Worst case scenario is the motor might burn out ( when I try it out just to address your concerns I will have numerous safety precautions in place) Am I being stupid because nobody seems to be able to say fairly precisely what the winch can pull,i don't know what else to do

 

[Rx7man]

Since it's hard to know exactly what the winch was designed for, I'd look at the cable it was supplied with... I know my winch is a 3/8th cable and is rated at 12,000 lb peak pull (empty drum), and it falls to about half that as the drum fills up... If you go by the breaking force of the cable, and take about 1/3rd of that, I think that would be a reasonable starting point for what the winch can pull with a full drum (there has to be some safety factor when the drum is empty, or the cable is less than new)...
a 2000kg boat/trailer on a 10* slope will be tan(10)*2000, or .176 * 2000 = 352kg, or about 800 lbs. It will of course take a little more to get it moving, but if you can guesstimate the capacity of the winch as being above 1000 lbs, it ought to do the job

 

[JBA]

I think it is best to focus on the 2000 kilo load you want to pull the 10° ramp and how that relates to the current cable rating of 2585 kilo rating assuming that cable has not already been upgraded from that originally specified for the unit. Using the 10° angle and the 2000 kilo load then the tension load on the cable (based entirely on the car plus boat and trailer weight, i.e. not taking into account the rolling friction of the trailer tires or the viscous drag of the water on the trailer and boat) = 2000 x sin 10° = 347 kilos which results in a Safety Factor = 2585/347 = 7.5 for your cable based upon its minimum breaking strength.

As for your 1 HP rated motor, based upon my prior calculation of a max load of 1199 kilos, the Hp required to provide 347 kilos of pull = 1 Hp x 347/1199 = .29 Hp with a Safety Factor of 3.5.

At the same time, this doesn't address the load limits of the gearbox and the chain drive, in particular; but, since this is a professionally designed unit I would expect those items to have similar Safety Factors. All of this based upon careful maintenance of the unit including the chain drive and wire cable in particular.

I hope this helps.

PS I was in the process of posting this when the above arrived.