Horsepower
A British unit of power, used to define the rate at which work is done. Has been replaced by the kilowatt, and an approximate conversion is 100hp = 75kW. Devised in the early days of steam engines it allowed a reasonable comparison for the layman with the rate of working for a good horse and was the result of an experiment by James Watt. One horsepower is the power needed to move 33,0001b one foot in one minute and equals 0.746 kilowatts. Brake horsepower (bhp) is calculated from the torque measured by brake test at the output shaft. Indicated horsepower (ihc), calculated from the mean effect pressure acting on the piston and taking into account the cross-sectional area of the bore, the length of stroke and the number of power strokes in a minute, is thus the work done on all pistons of an engine in one minute.
Calculating Horsepower
The question of horsepower is often a hotly debated issue. Sometimes an enthusiast is every bit as optimistic as their speedo. As often again a car enthusiast just does not know what their warmed-up engine develops and is wise enough not to make vague guesses. Settling an argument is not the only reason for wanting to know the true horse-power of a worked-over engine. The choice of gears, rear axle ratios and power-to-weight ratio (when a new chassis is being developed) all depend on the available sting under the bonnet.
In choosing gear ratios, you need to know not only the available b.h.p. but also the available torque. There are three ways of finding out the power facts of your engine. You can take it to a specialist firm who will test it on a dynamometer and draw up a complete power curve and torque graph. This is easily the best approach - but of course it is expensive and not always practical since only a few firms can afford to install a dynamometer. The second way is to determine accurately the acceleration, weight and top speed of the car and then calculate the horsepower necessary to provide such a performance. Thirdly, you can use a multiplication factor for each step you have taken in the warming up process and then estimate the power that the engine develops - or should develop if the job has been done correctly.
The Performance Method of Calculating Horsepower
The most popular method is the performance method, so we'll discuss that first. Simply applying a set of formulae is not enough. To do the job properly, it is necessary to understand where the figures come from. For this reason we must get quite clearly fixed in our minds the meaning of the words horsepower and torque. Torque is a turning thrust and is a product of a force times a distance. Horsepower, on the other hand, is the rate at which work is being done and there is a time factor involved. Examples are probably the best way of illustrating these points.
Take the case of a 13" wheel which is known to exert a force of 300 lbs. at the road surface when the car is accelerating. In this case the torque transmitted by the wheel is the force (300 lbs.) times the distance from the hub to the road surface (13"). This gives a figure of 3,900 lb. in. or 325 ft. lb. Suppose also that under the same conditions, the wheel is spinning round at 400 r.p.m. Then the horsepower can be calculated. First we have to find the work being done. Work is a product of distance times a force. The distance involved is 2 pi radians times the r.p.m., where a radian is the angle subtended at the centre of a circle by an arc of equal length as the radius of the circle. This is 2 pi x 400. (Pi is the accepted constant for circular measurement and equal to roughly 3.14).
We now know that work is being done at the rate of 2 pi x 400 x 325 lb. ft. per minute. This can easily be converted to horsepower, since by definition a horsepower is 550 ft. lb. per second. Thus to convert the above from ft lb. to b.h.p., we merely divide by 550 x 60. The horsepower at the wheel is (2 pi x 400 x 325 ) / (550 x 60) = (approx.) 24. Of course the horsepower available at the wheel is not the same as that developed by the engine. First, roughly eight per cent, of the engine power available at the clutch is lost in the gearbox and rear axle. In the indirect gears the power loss will be higher - probably about ten per cent. Thus the horsepower delivered at both wheels is roughly 90 percent, of the total engine b.h.p. Now the torque curve of an engine will depend entirely on the engine design. The manufacturers have a choice of providing plenty of torque well down the rev. scale (to give maximum top gear flexibility), or of providing it uniformly over a wide area of the rev range or of arranging so that its peak (in the power curve) almost coincides with maximum b.h.p.
We have already seen that horsepower is a product of torque times the distance. In this case it is a product of torque times r.p.m., divided by an appropriate conversion factor. The b.h.p. of an engine is low when the engine speed is low, but it rises with the r.p.m. until a peak point is reached. At this stage the drop in torque is exactly equal to the gain in b.h.p. produced by the increase in r.p.m. At higher engine speeds, the b.h.p. drops because the torque falls off. The aim when warming up an engine is to change its torque characteristics so that the peak torque point is reached as high up the r.p.m. scale as reasonably possible. In this event, maximum power is raised because, as we have already seen, it is a product of the r.p.m. and the torque.
The ideal is to bring about a condition where the peak of the torque curve is as close as possible to the peak of the b.h.p. curve. Then the torque multiplication between the engine and road wheels is at a maximum and the acceleration is at its liveliest. Of course, this makes the engine extremely inflexible and energetic use of the gearshift is essential. Sedan car manufacturers are concerned more with flexibility than sheer power. Often a large engine is deliberately equipped with restricted breathing, a small carburettor throat and a camshaft that encourages maximum torque at low r.p.m. This state of affairs makes the car ideal for the lazy motorist since he does not need to change out of top gear, except when starting from rest. It is also a wonderful plaything in the hands of the enthusiast, because it can be warmed up with most gratifying results.
With this knowledge digested, let's get back into gear and see how horsepower, torque and acceleration are connected. We need this knowledge to make the necessary calculations to assess the horsepower of a warmed up engine. Any student of dynamics knows that the law governing acceleration is P = Mf, where P is the force (expressed in poundals), M is the mass of the vehicle (in pounds) and f is the rate of acceleration. "P" is the one symbol that we r.eed to discuss. It is the accelerating force, that is the excess torque left over after the work has been done to overcome rolling resistance.
Colin Campbell in his excellent book "The Sports Car" released way back in the mid 1950s shed some very useful light on this subject. He said that at speeds up to 50 m.p.h., air and road resistances were small and for a normally designed sports car was around 80 lb. at 50 m.p.h. The next step in the calculation is best shown by a typical example given by Colin Campbell. He quotes the case of a sports car with a torque of 120 lb., a third overall gear ratio of 6:1 and a transmission efficiency of 90 per cent., with a tyre diameter of 27 in. In this case, the thrust at the rear wheels is (120 x 0.9 x 6 x 12) /13.5 = 575 lbs.
From this we must deduct the 80 lb. torque consumed by the rolling resistance. This gives us a net figure of 495 lb. f or acceleration. It can thus be seen that excess thrust is about 86 per cent, of the total engine torque. At low speeds it is even higher - possibly as much as 95 per cent. So if we consider the case of a car accelerating from 0-50, we can safely assume that an average of 90 per cent, of the total engine torque is available for acceleration. Let's go back to our formula: P = Mf. P - the accelerating force - is proportional to T, the torque of the engine. We can say then that F (the rate of acceleration) = K (T/M) where K is a constant. Thus the rate of acceleration is equal to KW/T where W is the weight of the car in cwt. and T is the total engine torque.
Colin Campbell calculated that K has a value of 50 for a well tuned sports car, when accelerated from 0 to 50 m.p.h. Thus Time (from 0-50 m.p.h) = 50W/T - or engine torque = 50W/Time. As an example, let's use the case of a 15 cwt. special taking seven seconds to accelerate from rest to an accurate 50 m.p.h. Then the total engine torque is (50 x 15)/7 = 107 ft. lb. Another formula gives the horsepower per laden ton for a vehicle, when the time is known, to accelerate through a standing quarter. The formula is: Time (in sec. for standing quarter) = 82 divided by the cube root of the horsepower per laden ton. Suppose that a warmed up sedan takes 20 seconds to cover the standing quarter mile. With an all-up weight of 22 cwt., its horsepower is not known. Then 20 = 82 / (cube root of P x 20 /22) or Cube root 10P/11 = 82/20 = 4.1.
This formula, by the way, is unusually accurate and the power obtained from it is invariably within five per cent, of the true figure. If you doubt it, try using the formula in conjunction with the known acceleration figures of cars of known weight and power. It is normal practice to estimate the power output of an engine in two or three different days and then take an average of the findings, so as to reduce any errors. A third way of determining power output is to calculate the true top speed of the car by timing it both ways over a measured quarter mile on the level. The aerodynamics of the car will greatly affect the operation of this formula but compensation factors can be applied as we shall see later.
The basics to know are that maximum speed varies with three things�the power available, the frontal area of the car and the shape of the body. Loosely, we can measure the frontal area of the car by multiplying its overall width by its overall height. It is measured (in this instance) in square inches, not feet, and we can call it "A". Then maximum speed = (K x BHP) / A. K - the constant - varies with the aerodynamics of the design. Our calculation, which are based on an elaborate formula devised by Colin Campbell, show that with an aerodyamic shape, such as the D-type Jaguar, K is equal to 2,000. For a medium design, such as the Porsche coupe, 2,500 is a closer figure. The Austin-Healey weighs in with a figure of about 3,000 and the M.G.T.D. (which is anything but streamlined) has K equalling 3,800.
From these figures, an enthusiast can estimate the value of K for their own car. Now, to apply the formula: suppose a home made special has an accurate top speed of 90 m.p.h., its overall dimensions are 60 inches wide by 45 inches high and the streamlining is poor so that K is assumed to be about 3,200. This is how we determine maximum b.h.p. V = (K x BHP) / area = (3,200 x BHP) / (45 x 60). Thus BHP = (27 x max speed) / 32 = approx 75. Yet another formula for determining maximum power from maximum speed is: Max. speed = 57 x cube root of BHP per square foot of frontal area. With the special we discussed before, let's apply the formula. Max speed = 57 x cube root of BHP 60 x 45 divided by 144 BHP i.e., 90 = 57 x cube root of BHP / 18.7. Thus 90 cubed/57 = BHP / 18.7 or approximately 1.7 cubed = BHP / 18.7 BHP = approx. 72 BHP.
Determining Horsepower Using The Multiplication Factor
A final means of estimating the horsepower of a warmed up engine is to use a multiplication factor for each modification made to the engine. This means that the stock b.h.p. (not S.A.E. horse-' power) for the engine must be known. The multiplication factors are themselves only estimates, but the overall figure works out reasonably accurate provided that the work is correctly done. If the intake breathing was considerably modified its effect on the power output would not be dramatic unless equivalent work were done to assist the exhaust.
Let's now find out the multiplication factor associated with a i-aised compression ratio. It won't help us much here to study the effects of a higher compression ratio, but it is enough to say that generally speaking, a power increase of seven per cent occurs when the ratio is raised by a full unit. That is, if a stock engine develops 60 b.h.p. with a 6.8 to 1 compression ratio, then it should develop 60 plus seven per cent, with a 7.8 to 1 ratio. The multiplication factor is thus 1.07 for each unit raise in compression ratio, unless detonation occurs.
Adding twin carburettors with a suitable manifold system provides a multiplication factor of 1.15 but this will rise to 1.20 if a third carburettor is fitted on a six cylinder unit. A semi-race camshaft gives a figure of 1.08 while a full race gives 1.12. Working over both the inlet and exhaust systems, together with larger valves and well polished porting gives 1.10. The figure obtained from oversized bores depends on the percentage increase in capacity. A five per cent, increase gives a multiplication factor of 1.035, a 10 per cent, increase gives 1.07, a 15 per cent, increase gives 1.105 and a 20 per cent, increase gives 1.175.
With a twin overhead camshaft conversion kit, properly designed and used in conjunction with most of the fore-going, the multiplication factor is 1.20. To estimate the power increase of an engine, each of these multiplication factors is multiplied with each other. The final figure is multiplied with the stock b.h.p. output of the engine. The result is the estimated power-output for the conversion. Let's suppose that a stock Holden engine develops 60 b.h.p. Its owner raises the compression ratio from 6.8 to 8.8. He adds a second carburettor with suitable manifolding, fits a semi-race camshaft and increases the bore until the capacity has been raised by 200 c.c.
The multiplication factors work out as follows: Compression ratio, 1.14; twin carburettors, 1.15; semi-race camshaft, 1.08. The extra 200 c.c. is roughly 10 per cent of the initial capacity and the multiplication figure is 1.07. Multiplied by each other, the figures give a sum of 1.51. The stock b.h.p. of 60 is now multiplied by 1.51 and the final figure is about 90 b.h.p. We've shown several different ways in which the power output of a warmed up engine can be calculated. For greatest accuracy, it is best to work out each method and take the mean of the four answers. But for normal use, any one of the above methods will give a reasonable accurate working figure.
