need to add this somewhere!!

my concern is NOT of making enough power, however its of making and lost of it at the wrong revs!! or should I of said torque!!! hehehehe!!!

so with the result of that, have ended up becoming even a bigger fan of google search!!

- Power and Torque -

ESSENTIAL CONCEPTS:

Torque is measured; Power is calculated

the above is very intersting

knew it as '''horse power is a by product of torque''

as a result of the little research that i did, I now know why and understand why a longer stroke = more torque!!

http://www.epi-eng.com/piston_engine_te ... torque.htm

In order to discuss powerplants in any depth, it is essential to understand the concepts of POWER and TORQUE.

HOWEVER, in order to understand POWER, you must first understand ENERGY and WORK. If you have not reviewed these concepts for a while, it would be helpful to do so before studying this article. CLICK HERE for a quick review of Energy and Work.

It often seems that people are confused about the relationship between POWER and TORQUE. For example, we have heard engine builders, camshaft consultants, and other technical experts ask customers:

"Do you want your engine to make HORSEPOWER or TORQUE?"

And the question is posed in a tone which strongly suggests that these experts believe power and torque are somehow mutually exclusive.

In fact, the opposite is true, and you should be clear on these facts:

1. POWER (the rate of doing WORK) is dependent on TORQUE and RPM.

2. TORQUE and RPM are the MEASURED quantities of engine output.

3. POWER is CALCULATED from torque and RPM, by the following equation:

HP = Torque x RPM ÷ 5252

(At the bottom of this page, the derivation of that equation is shown, for anyone interested.)

An engine produces POWER by providing a ROTATING SHAFT which can exert a given amount of TORQUE on a load at a given RPM. The amount of TORQUE the engine can exert usually varies with RPM.

A dynamometer determines the POWER an engine produces by applying a load to the engine output shaft by means of a water brake, a generator, an eddy-current absorber, or any other controllable device capable of absorbing power. The dynamometer control system causes the absorber to exactly match the amount of torque the engine is producing at that instant, then measures that TORQUE as well as the RPM of the engine shaft, and from those two measurements, it calculates observed power. Then it applies various factors (air temperature, barometric pressure, relative humidity) in order to correct the observed power to the value it would have been if it had been measured at standard atmospheric conditions (corrected power).

TORQUE

TORQUE is defined as a FORCE around a given point, applied at a RADIUS from that point. Note that the unit of TORQUE is one pound-foot (often misstated), while the unit of WORK is one foot-pound.

Referring to Figure 1, assume that the handle is attached to the crank-arm so that it is parallel to the supported shaft and is located at a radius of 12" from the center of the shaft. In this example, consider the shaft to be fixed to the wall. Let the arrow represent a 100 lb. force, applied in a direction perpendicular to both the handle and the crank-arm, as shown.

Because the shaft is fixed to the wall, the shaft does not turn, but there is a torque of 100 pounds-feet (100 pounds times 1 foot) applied to the shaft.

Note that if the crank-arm in the sketch was twice as long (i.e. the handle was located 24" from the center of the shaft), the same 100 pound force applied to the handle would produce 200 lb-ft of torque (100 pounds times 2 feet).

POWER

POWER is the measure of how much WORK can be done in a specified TIME. In the example on the Work and Energy page, the guy pushing the car did 16,500 foot-pounds of WORK. If he did that work in two minutes, he would have produced 8250 foot-pounds per minute of POWER (165 feet x 100 pounds ÷ 2 minutes). If you are unclear about WORK and ENERGY, it would be a benefit to review those concepts HERE.

In the same way that one ton is a large amount of weight (by definition, 2000 pounds), one horsepower is a large amount of power. The definition of one horsepower is 33,000 foot-pounds per minute. The power which the guy produced by pushing his car across the lot (8250 foot-pounds-per-minute) equals ¼ horsepower (8,250 ÷ 33,000).

OK, all that’s fine, but how does pushing a car across a parking lot relate to rotating machinery?

Consider the following change to the handle-and-crank-arm sketch above. The handle is still 12" from the center of the shaft, but now, instead of being fixed to the wall, the shaft now goes through the wall, supported by frictionless bearings, and is attached to a generator behind the wall.

Suppose, as illustrated in Figure 2, that a constant force of 100 lbs. is somehow applied to the handle so that the force is always perpendicular to both the handle and the crank-arm as the crank turns. In other words, the "arrow" rotates with the handle and remains in the same position relative to the crank and handle, as shown in the sequence below. (That is called a "tangential force").

If that constant 100 lb. tangential force applied to the 12" handle (100 lb-ft of torque) causes the shaft to rotate at 2000 RPM, then the power the shaft is transmitting to the generator behind the wall is 38 HP, calculated as follows:

100 lb-ft of torque (100 lb. x 1 foot) times 2000 RPM divided by 5252 is 38 HP.

The following examples illustrate several different values of TORQUE which produce 300 HP.

Example 1: How much TORQUE is required to produce 300 HP at 2700 RPM?

since HP = TORQUE x RPM ÷ 5252

then by rearranging the equation:

TORQUE = HP x 5252 ÷ RPM

Answer: TORQUE = 300 x 5252 ÷ 2700 = 584 lb-ft.

Example 2: How much TORQUE is required to produce 300 HP at 4600 RPM?

Answer: TORQUE = 300 x 5252 ÷ 4600 = 343 lb-ft.

Example 3: How much TORQUE is required to produce 300 HP at 8000 RPM?

Answer: TORQUE = 300 x 5252 ÷ 8000 = 197 lb-ft.

Example 4: How much TORQUE does the 41,000 RPM turbine section of a 300 HP gas turbine engine produce?

Answer: TORQUE = 300 x 5252 ÷ 41,000 = 38.4 lb-ft.

Example 5: The output shaft of the gearbox of the engine in Example 4 above turns at 1591 RPM. How much TORQUE is available on that shaft?

Answer: TORQUE = 300 x 5252 ÷ 1591 = 991 lb-ft.

(ignoring losses in the gearbox, of course).

The point to be taken from those numbers is that a given amount of horsepower can be made from an infinite number of combinations of torque and RPM.

Think of it another way: In cars of equal weight, a 2-liter twin-cam engine that makes 300 HP at 8000 RPM and 400 HP at 10,000 RPM will get you out of a corner just as well as a 5-liter engine that makes 300 HP at 4000 RPM and 400 HP at 5000 RPM. (In fact, in cars of equal weight, the smaller engine will probably race BETTER because it's much lighter, therefore puts less weight on the front end.)

there is a lot more info, however this explains a lot!!

the link is at the beginning of the post!!