Effective winching requires an understanding of the mechanics involved in winching.
There are four main factors which influence resistance of motion when winching. These
are:
At the end of this discussion are two examples, one for
winching
a Jeep up a sand dune
, and another for
winching
a Hummer out of clinging clay
.
Vehcile Momevement Resitance
This is influenced by the state of the tyres (inflated/deflated), friction in
the drive-train, and vehicle weight. Winching is easier on normally-inflated tyres
as flat tyres add additional drag.
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Vehicle Weight
This refers to the gross vehicle mass, which includes passengers, baggage, fuel
load and anything else aboard the vehicle when winching takes place.
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The Surface to be Traversed
This is the largest variable when winching. A vehicle in good running condition
on a hard, level tarmac surface will only require a force of about 4% of its total weight
to induce motion. A vehicle to be recovered from a bog will, however, require a pull
equivalent to about 50% of the total weight of the vehicle. The table in the sidebar shows that
different kinds of surfaces require proportionate efforts to induce vehicle motion.
A simple formula can be used to show that approximate rolling resistance
of a vehicle in good condition on a flat surface can be predicted.
Pulling Effort required (in lbs) |
|
= |
|
Weight of Vehicle |
Resistance Co-efficient |
Using this formula in the example that follows, we calculate the effort required
to move a vehicle weighing about 4500lbs along a flat beach of hard, wet sand (which
has a resistance factor of 1/6):
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Surface Gradient
Obviously not all surfaces are flat, and the steeper the surface, the greater the
effort required to move the vehicle. The calculation must therefore include the
gradient resistance co-efficient.
Typically, the gradient will only be of a short distance, such as a ditch or rock,
or it may be of a longer distance, such as a long climb up a hill. Even for a relatively
short pull, gradient resistance must be taken into account. For practical purposes,
gradient resistance can be taken as 1/60th of the weight of the vehicle for each degree
of the slope, up to a 45-degree incline.
Gradient Resistance |
|
= |
|
Gradient x Vehicle Weight |
60 |
Based on this formula, on a 15-degree slope, gradient resitance will be 15/60 of the
weight of the vehicle, which is 1/4 of the total vehicle weight. That the slope to be
negotiated to all intents and purposes is only 1ft high makes no difference to the
calculations, and should be borne in mind when pulling vehicles over ridges. Consequently,
if we combine the weight of the vehicle, the type of surface to be traversed and the
gradient to be overcome we need to use the following calculation:
Pulling Effort required (in lbs) |
|
= |
|
Weight of Vehicle |
|
+ |
|
Gradient |
|
x |
|
Vehicle Weight |
Surface to be traversed |
60 |
The final winching formula therefore looks like this:
Pulling Effort required (in lbs) |
|
= |
|
W |
|
+ |
|
(G x W) |
|
S |
60 |
Where |
|
W |
|
= |
|
Total vehicle weight |
S |
= |
Surface to be traversed |
G |
= |
Angle of gradient (in degrees) |
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Example 1
In the first example we'll winch an unladen Jeep Wrangler TJ up a sand dune
of dry, loose sand with an incline of 15 degrees. Using the winching formula
from above, we get the following:
W |
|
= |
|
3383 lb. (unladen weight of a TJ) |
S |
= |
1/4 (resitance factor of surface to be traversed) |
G |
= |
15 (angle of gradient in degrees) |
Using these figures, we now have:
Pulling Effort required |
|
= |
|
3383 |
|
+ |
|
(15 x 3383) |
4 |
60 |
|
|
|
|
|
= |
3383 |
+ |
846 |
4 |
|
|
|
|
= |
846 |
+ |
846 |
|
|
|
|
= |
1692 Lb. |
|
|
So in this case 1692 Lb. of effort is required to pull an unladen Jeep Wrangler up
a 15 degree slope covered with soft, dry sand.
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Example 2
In the second example we'll winch an unladen Hummer up a 32-degree slope of clinging
clay. Again using the winching formula from above, we get the following:
W |
|
= |
|
6814 lb. (unladen weight of an open-top Hummer) |
S |
= |
1/2 (resitance factor of surface to be traversed) |
G |
= |
32 (angle of gradient in degrees) |
Using these figures, we now have:
Pulling Effort required |
|
= |
|
6814 |
|
+ |
|
(32 x 6814) |
2 |
60 |
|
|
|
|
|
= |
6814 |
+ |
3634 |
4 |
|
|
|
|
= |
3407 |
+ |
3634 |
|
|
|
|
= |
7041 Lb. |
|
|
7041 Lb. of effort is required to winch the Hummer out of the clay. As you can see,
the winching effort increases exponentially with the severity of the gradient.
|