Horizontal pulling capacity of a vertical hoist

[SirDick Diodenob esquire]

I have a cheap chinese made hoist which is rated at 200kG or 400kG using a pulley, I want to use it to pull horizontally a car weighing about 1600kG , not quite horizontally but over a slight incline and a small kerb. Whats the difference between a vertical hoist and a 12V winch , have seen Longhorne winches etc for a round £30 to £50 on ebay rated at 2000lbs from stationary to 6000lbs moving, I dont want to buy yet another item I will not use much if the vertical hoist will do.

 

[dan.taylor.1]

Vertical hoists are designed to hold the load on a brake or catch, winches don't normally.

Dan

 

[AndyStobbs]

Weight of the car isn't really that relavant. How well does it roll is the question?

 

[SirDick Diodenob esquire]

I hadnt thought of that but its something that would help , main concern is would 200kg of vertical lifting capacity be capable of pulling a 1600kg car. The motor on the hoist is 1kW and fairly low geared . I am not too bothered about overloading it but I need to make up a jig with poles in the ground to attach it to so some work involved so wanting to have an idea if it will have the pulling power before spending time welding up a frame.

The car rolls easy on a flat road but its on a slight incline maybe 5 to 6 degrees and then a small lowered kerb about 2inches . We have pushed cars in the past and its taken three of us to get it up the same kerb/incline but as you get older pushing cars is bad for your life expectancy so need to find another way.

 

[BrokenBiker]

can you push the car, on your own just to get it rolling? If yes, even just a slight bit then your hoist should do it. Its rolling weight you need to consider, not gross weight

 

[AndyStobbs]

Just imagine your car sat on the edge of a cliff, with the rope passing over the end of the cliff on a pulley, cliff is a slope 5 or 6 degrees like your drive. If you attach 200kg to the end of the rope and hang it over the cliff would that be enough? Quite possibly not until the car was moving, double it to 400kg, that should do it, again might help to give it a little shove to get it moving.

Use a pulley on the car and anchor the free end of the rope to your frame.

 

[AndyStobbs]

Oh and I think getting older generally is bad for your life expectancy to be honest - and who wants to arrive at the pearly gates in a good body? Planning to wear mine out myself.

 

[CKJDJK]

Here is the calculations recovery folk use to work out how strong a winch is required and the tackle required.

SECTION 3 ‑ ESTIMATING THE FORCE REQUIRED
a. Rolling Resistance (RR).
b. Gradient Resistance (GR).
c. Damage Resistance (DR).
d. Safety Factor (SF).
The sum of these resistances will amount to the EP (Estimated Pull).
ROLLING RESISTANCE (RR)
Type of Ground Ground Factor
Smooth road 25
Grass 7
Hard wet sand 6
Gravel and soft wet sand 5
Loose dry sand 4
Shingle beach 3
Soft clay and black mud 2
520. The degree of resistance offered by the ground on which a casualty rests is determined by:
a. The weight of the vehicle (WOV).
b. The type of ground.
RR = WOV where RR = Rolling Resistance
GF WOV = Weight of Vehicle
GF=Ground Factor
Example: Weight of vehicle =14 tonnes
Type of ground = deep bog (GF 2)
Therefore: RR = 14 = 7 tonnes
2
GRADIENT RESISTANCE (GR)
a. The weight of the vehicle.
b. The gradient or degree of slope (DOS).
For gradients up to 45 degrees gradient resistance is taken as proportional to 1/60th of the weight of the vehicle. Thus:
GR = WOV X DOS
60
Example: With a 52 Tonne MBT
On a 20° slope ` GR=WOVx20° = 52x20 =17.3t
60° 60°
Rounded up = 18t
On a 30° slope GR = WOV x 30° = 25x 30°= 26t
60°60°
On a 45" slope GR = WOV x 45° = 52 x 45° = 39t
60° 60°
For a gradient greater than 45 degrees gradient resistance is taken as equal to the weight of the vehicle. It will be noted that the resultant figures are rounded up to nearest whole number. This is normal in all recovery calculations and will be seen to be applied in the examples that follow.
523. One method of assessing the degree of slope can be achieved by drawing a square on a sheet of paper and then bisecting it. This first bisection will give an angle of 45° and subsequently if that was bisected would give an angle of approximately 30°. If this angle was again bisected it would then give an angle of approximately 15°. This sheet of paper could then be held at arms length and the lines drawn can be compared against the slope in question, from this a prediction could be made on the degree of slope. This square is shown in the diagram Fig 11 below.

DAMAGE RESISTANCE (DR)
524. In recovery, the term damage resistance means the resistance arising from locked tracks or wheels. The degree of damage resistance depends on:
a. Weight of vehicle.
b. The number of locked wheels or tracks.
525. For estimating purposes, the damage resistance offered by locked tracks may be as much as:
a. One locked track =1/3 WOV tonnes
b. Two locked tracks = 2/3 WOV tonnes
For wheeled vehicles the resistance is calculated by:
a. Half or less the wheels/axles locked =1/3 WOV tonnes
b. More than half the wheels/axles locked = 2/3 WOV tonnes
526. In the following two examples, taking the weight of the vehicle in each as 14 tonnes, the damage resistance figures would be:
a. Two locked tracks DR = 2 x 14= 10 tonnes
3
b. Half the wheels locked DR= 14 = 5 tonnes
3
SAFETY FACTOR (SF)
527.A safety factor is involved in the overall calculation to allow for:
a. Worn equipment.
b. Unforeseen circumstances.
c. Miscalculations.
d. Frictional loss.
The safety factor is applied when completing the estimation of the total force to be employed. The safety factor figure is taken as 1/4 or 25% of the sum of rolling resistance, gradient resistance and damage resistance, thus the formula is:
RR + GR + DR
4
For example:
If RR = 7 tonnes
GR = 3 tonnes
DR = 5 tonnes
7+3+5 = 15 = 4 tonnes
4 4
EXAMPLE OF A CALCULATION
528.The estimates of the four component factors (see Para 518) work out, in a particular
case, as:
RR = 7 tonnes: GR = 3 tonnes: DR = 5 tonnes: SF = 4 tonnes
The estimated total force or pull is therefore:
7+3+5+4=19tonnes

 

[Robotstar5]

I reckon you'd need about 250Kg line pull for your car.

 

[fizzy]

Well I am a great fan of tifors.
I even have the smallest one. The Jockey hoist which is rated at 300kgs lifting. It is great carry it in an aluminium briefcase type tool box like James Bond! It is amazing how powerful it is. Once used it to winch an American Limo up a very steep drive. As an anchor point we had a Nissan Bluebird. In gear and handbrake on it was dragging the car. We chocked the wheels with bricks - it was pulling the car over the bricks - amazing. also great for moving lathes and milling machines etc. Easier to handle than the full size tirfors!

 

[Wendelspanswick]

The biggest problem with Tirfors is the price tag!

 

[SirDick Diodenob esquire]

Thanks for all the advice. The electric winch using a single wire rope rated at 200kg would not move the car (the 1.5hp motor stalled and would have burned out ) , using one pulley double looped giving 400kg was more than enough so I reckon that the 250kg stated by Robotstar is about right . Also tried using a 4000kg manual ratchet winch ( cheap chinese costing £15 ) , this also moved it easily but takes a long time as it only moves the car about 5 feet each time then have to start again. Never heard of a Tirfor before but having had a look on ebay they cost more than I can afford but wouldnt mind getting one for christmas.

 

[norlander]

check out your hire shops for tirfors.