in which P is the atmospheric pressure; P1 is the final pressure, and V/V1 is the compression ratio, therefore P1 = 14.7 (4.5)1.3 = 104 pounds per square inch, absolute.
“That is, 104 pounds per square inch is the most efficient final compression pressure to have for this engine at sea level, since it comes directly from the compression ratio.
“Now supposing we consider that the altitude is 7,000 feet above sea level. At this height the atmospheric pressure is 11.25 pounds per square inch, approximately. In this case we can again substitute in the formula, using the new atmospheric pressure figure. The equation becomes:
P1 = 11.25 (4.5)1.3—79.4 pounds per square inch, absolute.
“Therefore we now have a final compression pressure of only 79.4 pounds per square inch, which is considerably below the pressure we have just found to be the most efficient for the motor. The resulting power drop is evident.
“It should be borne in mind that these final compression pressures are absolute pressures—that is, they include the atmospheric pressure. In the first case, to get the pressure above atmospheric you would subtract 14.7 and in the latter 11.25 would have to be deducted. In other words, where the sea level compression is 89.3 pounds per square inch above the atmosphere, the same motor will have only a compression pressure of 68.15 pounds per square inch above the atmosphere at 7,000 feet elevation.
“From the above it is evident that in order to bring the final compression pressure up to the efficient figure we have determined, a different compression ratio would have to be used. That is, the final volume would have to be less, and as it is impossible to vary this to meet the conditions of altitude, the loss of power cannot be helped except by the replacing of the standard pistons with some that are longer above the wrist-pin so as to reduce the space above the pistons when on top center. Then if the ratio is thereby raised to some such figures as 5 to 1, the engine will again have its proper final pressure, but it will still not have as much power as it would have at sea level, since the horse-power varies directly with the atmospheric pressure, final compression being kept constant. That is, at 7,000 feet the horse-power of an engine that had 40 horse-power at sea level would be equal to
| 11.25 | = 30.6 horse-power. |
| 14.7 |
“If the original compression ratio of 4.5 were retained, the drop in horse-power would be even greater than this. These computations and remarks will make it clear that the designer who contemplates building an airplane for high altitude use should see to it that it is of sufficient power to compensate for the drop that is inevitable when it is up in the air. This is often illustrated in stationary gas-engine installations. An engine that had a sea-level rating amply sufficient for the work required, might not be powerful enough when brought up several thousand feet.” When one considers that airplanes attain heights of over 18,000 feet, it will be evident that an ample margin of engine power is necessary.