Simple sums on braking road transfer
Page 47
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OF ALL the phenomena relating to braking performance, load transfer during an emergency stop has assumed the greatest importance in recent years. The mathematics of load transfer are fairly simple and it is useful to do one or two sums to show how much load is transferred in a typical case with variations in the height of the centre of gravity and braking efficiency.
The example is taken of a near-maximum load four-wheeler (with a gross weight of 15 tons) having a relatively short wheelbase of 15ft; the shorter the wheelbase, the greater is the load transfer, other things being equal. The static load on the front axle is 5 tons and the rear axle load is 10 tons, and from this it is possible to calculate the distances from the axles of the perpendicular on which the centre of gravity lies. It should be noted that load transfer is not a function of the position of the c. of g. in the horizontal plane, except in so far as it affects the loads carried by the axles.
As shown by the sums, the centre of gravity lies on a perpendicular that is 10ft from the front axle and 5ft from the rear axle. If the c. of g. height is 511 and the braking efficiency is 50 per cent, the load transferred from the front to the rear axle is 2+ tons.
The position of the c. of g. and the load transfer value when the vehicle is braked can be calculated in the following ways. Reference should be made to the diagram.
Taking moments about the rearaxle: Front axle load x wheelbase = total weight of vehicle (acting through centre of gravity) x distance of the vertical line on which the c. of g. lies from the centre line of the back axle(b).
.•• 5 x 15 = 15 x b b = —75 = 5 ft.
As a check, moments can be taken about front axle to find the distance of the c. of g. line from the axle.
10 x 15 = 15 x a a = 10ft.
Braking force F = brake efficiency (percentage of g) x weight of vehicle. If
efficiency = 50 per cent, F = —50 x 15 =
tons. F x height of centre of gravity (h) — load transfer L (weight added to the front axle and reduction of weight on the rear axle) x wheelbase..'. FL = Ld.
If L = 5ft 7.5 x 5 = Lx 15 .'• L = 2.5 tons. .'. weight on front axle = 5 + 2.5 = 7.5 tons and weight on back axle = 10 — 2.5 = 7.5 tons.
Using another formula to check: Weight transfer = braking force x wheelbase (L= Fh wheelbase ) 7.5 x 5 •• L = = 2.5 tons.
15 Taking the extreme (but not impossible) case in which the height of the centre of gravity = 10ft and braking efficiency = 100 per cent: Braking force = weight of vehicle = 15 tons, and weight transfer — 15 x 10 15 = 10 tons. The load on the front axle would then be 15 tons and the load on the rear axle would be reduced to 5 tons.
If the 15ton vehicle had a wheelbase of 20ft and a centre of gravity of 5ft, the load transfer would be 1.875 tons if the vehicle were braked with an efficiency of 50 per cent. The front axle load would be increased to 6.875 tons, while the rear axle load would be reduced to 8.125 tons.
In the case of a vehicle that is braked on a slope, the rate of retardation equals the braking efficiency plus the percentage gradient if the vehicle is going uphill and minus the percentage if it is travelling downhill. Given that the gradient was 1 in 5(20 per cent) and the brakes were applied with a force that would give 50 per cent efficiency on the level, the rate of retardation (overall braking efficiency) uphill would be 70 per cent; whereas the efficiency downhill would be 30 per cent.
When considering brake load transfer of a vehicle on a slope it should be noted that load is transferred from front to rear if the vehicle is going uphill at a steady speed or is held stationary on the hill. Taking the example of a 20 per cent slope, load transfer would be approximately (but not exactly) equal to that produced if the vehicle were braked with an efficiency of 20 per cent when it was travelling backwards on the level. If the brakes were applied (when the vehicle was travelling up the slope) with an efficiency of 50 per cent, load transfer from the rear wheels to the front wheels compared with the static loads would be equivalent to transfer derived from braking on the level with an efficiency of 30 per cent.
By the same tokens, the load on the front wheels of a vehicle on a 20 per cent slope facing downwards would be approximately equal to the transfer when braking on the level with an efficiency of 20 per cent and, if a 50 per cent braking force were applied, the load would be increased to an amount equal to that produced by a 70 per cent stop on the level.
In practice, of course, a heavy vehicle is rarely braked to give a high rate of retardation on a very steep slope such as 1 in 5 but travels down the slope at a steady speed with a lower ratio engaged to augment the power of the brakes. To cite a case in which a vehicle is retarded at a rate equivalent to an on-the-level efficiency of 50 per cent is of academic interest only. Examining the performance of vehicles on steep gradients is, therefore, more important with regard to braking power than load transfer.
An articulated vehicle may be considered as two vehicles when assessments are made of braking load transfer, the front wheels of the second vehicle (the semi-trailer) being the rear wheels of the tractive unit. Load-transfer calculations can be done in the way indicated with the qualification that allowance has to be made for the position of the coupling king pin. If this is located vertically above the axle of the tractive unit all the weight transferred will be added to the load on the axle.
It is now normal practice to locate the kingpin ahead of the driving axle to transfer some of the static weight imposed by the coupling to the front axle, and the percentage of the braking load transferred to the axle is equal to the proportion of the static load transferred. The percentage is equal to the load multiplied by the distance of the king pin from the axle centre line, divided by the wheelbase. If the king-pin-to-axle distance was lft and the wheelbase of the tractive unit were 10ft, hth of the braking load would be transferred to the front axle.