The mechanical advantage of the wheel in surmounting an obstacle may be computed from the principle of the lever.
Let the wheel, Fig. 2, touch the horizontal line of traction in the point A and meet a protuberance B D. Suppose the line of draught C P to be parallel to A B. Join C D and draw the perpen diculars D E and D F. We may suppose the power to be applied at E and the weight at F, and the action is then the same as the bent lever E D F turning round the fulcrum at D. Hence P: W :: F D: D E.
But FD:DE::tanFCD:1, and tan F C D = tan 2 (D A B); therefore P = W tan 2 (D A B). Now it is obvious that the angle D A B increases as the radius of the circle diminishes; and therefore; the weight W being constant, the power required to overcome an obstacle of given height is diminished when the diameter is increased. Large wheels are therefore the best adapted for surmounting inequalities of the road.
There are, however, circumstances which provide limits to the height of the wheels of vehicles. If the radius A C exceeds the height of that part of the horse to which the traces are attached, the line of traction C P will be inclined to the horse, and part of the power will be exerted in pressing the wheel against the ground. The best average size of wheels is considered to be about 6 feet in diameter.
Wheels of large diameter do less damage to a road than small ones, and cause less draught for the horses.
With the same load, a two-wheeled cart does far more damage than one with four wheels, and this because of their sudden and irregular twisting motion in the trackway.
Grade Resistance is due to the action of gravity, and is the same on good and bad roads. On level roads its effect is immaterial, as it acts in a direction perpendicular to the plane of the horizon, and neither accelerates nor retards motion. On inclined roads it offers considerable resistance, proportional to the steepness of the incline.
The resistance due to gravity on any incline in pounds per ton is equal to The following table shows the resistance due to gravity on dif ferent grades.
Resistance Due to Gravity on Different Inclinations.
Grade 1 in 20 30 40 50 60 70 80 90 100 200 300 400 Rise in feet per mile 264 176 132 105 88 75 66 58 52 26 17 13 Resistance in lb. per ton 112 74+ 56 45 38 32 28 25 2211+ 71 5+ The additional resistance caused by inclines may be investigated in the following manner: Suppose the whole weight to be borne on one pair of wheels, and that the tractive force is applied in a direction parallel to the surface of the road.
Let A B in Fig. 3 represent a portion of the inclined road, C being a vehicle just sustained in its position by a force acting in the direction C D. It is evident that the vehicle is kept in its position by three forces; namely, by its own weight W acting in the vertical direction C F, by the force F applied in the direction C D parallel to the surface of the road, and by the pressure P which the vehicle exerts against the surface of the road acting in the direction C E perpendicular to same. To determine the relative tude of these three forces, draw a horizontal line A G and the vertical one B G; then, since the two lines C F and B G are parallel and are both cut by the line A B, they must make the two angles C F E and A B G equal; also the two angles C E F and A G B are equal; therefore, the remaining angles F C E and B A G are equal, and the two triangles C F E and A B G are similar. And as the three sides of the former are proportional to the three forces by which the vehicle is sustained, so also are the three sides of the latter; namely, A B or the length of the road is proportional to W, or the weight of the vehicle; B G, or the vertical rise in the same, to F, or the force required to sustain the vehicle on the incline; and A G, or the horizontal distance in which the iise occurs, to P, or the force with which the vehicle presses upon the surface of the road. Therefore, and If to A G such a value be assigned that the vertical rise of the road is exactly one foot, then in which A is the angle B A G.