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										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) |   Back to top
 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.
									  Back to top
 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.
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