Friday, June 23, 2017 | TSTM

Horse Power

Horsepower is generally used to relate how powerful an engine is in relation to the number of horses it would take to equal it's power and is generally stated to be "33,000 foot pounds of work per minute". The horse power output of an engine varies depending on the torque and the RPM of the crankshaft. The formula for computing horse power is HP = ((Torque)*(RPM))/5252. The 5252 is a constant that represents 1 horse power, if Torque (foot pounds) * RPM is 5252 then you have 1 horse power generated as output. This constant is calculated by taking the amount of foot pounds in 1 horse power (33,000) and converting it into radians per revolution (2*PI) so 5252 = 33000/(2*PI). The general equation is "Power = (Force * Distance) / Time" which in this case force is torque in foot pounds, distance is the revolutions per minute of the crank shaft, time is calculated as 1 horsepower in a single revolution and power is in units of "horse power" or 33,000 foot pounds.

Let's take a look at severals specifications for the KTM 495's maximum power output.

  • 1984 KTM 495 = 50 HP @ 6400 RPM
  • 1982 KTM 495 = 56 HP @ 6400 RPM
  • 1981 KTM 495 = 53 HP @ 7000 RPM

We can then figure out what the torque for the engine is at peak power based on the above figures (NOTE: There are various methods to calculate horse power that are described in Wikipedia).

  • HP = ((Torque)*(RPM))/5252
  • HP*5252 = (((Torque)*(RPM))/5252)*5252
  • (HP*5252)/RPM = (Torque*RPM)/RPM
  • (HP*5252)/RPM = Torque
  • 1984 KTM 495 (50*5252)/6400=41
  • 1982 KTM 495 (56*5252)/6400=46
  • 1981 KTM 495 (53*5252)/7000=40

Torque is what pressure the combustion on the piston puts onto the crank in which the connecting rod rotates the crank shaft itself. This pressure is what causes the crankshaft to rotate and generate "power" or "horse power" as output. There is a peak at some RPM of the crank which is the maximum torque is achieved by the combustion engine and from there as the RPMs rev higher starts to drop off. Using the formula HP = (T*RPM)/5252 you can see that at a certain RPM after this you reach "Peak Power" which is the maximum possible power output. This is the PowerBand of the engine; the RPMs from peak torque to peak power. It is not possible for Peak Torque to come after peak power besides the physical description the math formula does not make this possible. I.E. if both Torque and RPMs are increasing there is no way for the formula HP = (T*RPM)/5252 to generate a smaller number and as such you would not have reached "Peak Power".

Let's take the above example of the 84 KTM 495. Torque = 41; RPM = 6400; HP = 50. 6500 RPM must be less than to equal to 50 HP since peak power was already achieved at 6400 RPM which means that Torque must be decreasing. If torque remains the same at 6600 RPMs you gain a horse power and a half (41*6600)/5252 = 51.5 HP. This means torque must be decreasing however by how much is hard to say. There are tools that are used to measure the torque graph output of an engine such as a dynamo. There are likely mathematical methods that could be used provided you had a lot of details about the engine such as the force of the combustion, the length of the connecting rod, the friction in the bearings, the weight of the crank shaft, the engine timing, etc. and could calculate a model based on this data.

The next question is let's say that the 1982 KTM 495 is 56 HP @ 6400 RPM with 46 Torque at the crank. What does this mean for the rear wheel? Well, the KTM 495 has a primary drive ratio of 1:2.5, first gear is also 1:2.5 and based on a 52 tooth rear sprocket and 14 tooth front sprocket you have a ratio of 1:3.71. The first thing to note is that Torque is multiplyed by the gear ratios so 46 Torque * 2.5*2.5*3.71 = 1066.6 Torque at the rear wheel. This is how you can calculate how much Torque you end up losing or gaining based on your sprocket setup.

The reason though that when you change the sprockets power seems to increase or decrease is because of how much torque is applied at the rear wheel. Changing sprockets could increase or decrease torque. Let's change the rear sprocket of the KTM now to 40 tooth sprocket. 40/14 = 1:2.85 ratio. 46 Torque * 2.5*2.5*2.85 = 819.4 Torque at the rear wheel. This is ~246 less torque and let's calculate both RPMs as follows 6400/2.5/2.5/2.85=359 and 6400/2.5/2.5/3.71 = 276. 276*1066/5252 = 56 HP and 359*819/5252 = 56HP. This is obvious since RPMs are multipled and Torque is divided by the same numbers that horse power would remain the same since division and multiplication cancel each other out. The problem is that you must spin the wheel faster with less torque which means its harder to reach that RPM.

The rear wheel needs to spin after all and if you can't get it to spin you can't achieve horse power and you can't achieve that RPM. The easier it is for torque to be high at the rear wheel when the wheel is turning slower will make it easier to get the bike into motion. Also due to friction, etc. there is likely some horse power lost and in general specification ratings listed a lot of times do not reflect actual horse power since it seems unlikely that it would be a constant. Engines are affected by wear, friction, heat/cold, weight, spark, etc.

However, there is a difference between Horse Power at the "Rear Wheel" verse "Horse Power" read at the crank. There are losses of horse power from the crank and the result of the real work being done is termed "Rear Wheel Horse Power" or horse power used in actual work. This is generally less than the horse power specifications quoted for the engine unless specifically stated that they are using rear wheel horse power. There is also the question of added corrections to the readings. There is an article in the references which discusses the sales of dynos which 'correct' the output resulting numbers.

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