Rick Carter

I had this saved and forgot about it until recent postings jogged my memory; click the link to use the calculator.

http://home.austin.rr.com/turbolexus..._Dyno_Calc.htm

This calculator determines how much the air conditions on any given day will alter the horsepower of a forced induction internal combustion engine. The Forced Induction Dyno Horsepower Corrector determines the relative horsepower available at any temperature, barometric pressure, dew point, and altitude. The calculations are derived from SAE J1349 Revision JUN90 and modified for the special needs of forced induction.

To use this calculator, just enter the air temperature, barometric pressure, dew point, altitude, and gauge boost level. Then click on the calculate button.


A short example should illustrate why this is necessary:

For simplicity's sake, we will leave out temperature and humidity and assume, for this example, they are equal at the sea level and altitude conditions we are testing at. This leaves the only basic variable causing differences in power levels, between sea level and altitude, the absolute air pressure (atmospheric pressure + boost). Rather, how many molecules of air occupy a given space, which is directly proportional to the pressure per square inch. The more air ingested, the more power that should be made.

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We will test 2 engines, one natural aspirated and one forced induction, each at 2 locations, both sea level(0ft) and at 5000ft

The naturally aspirated engine at sea level(0ft) will ingest 14.7 psi(atm_0ft) (atmospheric air pressure) and makes 100hp

The naturally aspirated engine at 5000ft will ingest 12.2 psi(atm_5000ft)

At 5000ft the naturally aspirated engine makes (12.2/14.7) * 100hp = 83.0 hp


This results in a needed correction factor of 1.20 for a naturally aspirated engine at 5000ft

Now, the forced induction engine running 10 psi(g) (gauge indicated pressure) at sea level will ingest 14.7 psi(atm_0ft) + 10psi(g) = 24.7 atm(0ft_10psi) [total air pressure] and makes 100hp

The forced induction engine running 10 psi(g) at 5000ft will ingest 12.2 psi(atm_5000ft) + 10 psi(g) = 22.2 (5000ft_10psi)

At 5000ft the forced induction engine makes (22.2/24.7) * 100hp = 89.9 hp

This results in a needed correction factor of 1.11 for a forced induction engine at 5000ft

Now this doesnt take into account the SAE 1.18 multiplication and .18 subtraction factors which are included in their formula, but those are only in the SAE formula as a result of emperical data/testing and have no effect on this illustration and only result in an even higher correction factor for both engine types. Other factors can also affect the necessary corrections, but are not part of the formula. One of them is the efficiency of the turbos, which change at altitude as the pressure ratio for a given boost level varies with altitude and temp. Unfortunately including such factors would greatly increase the information needed (pulling info from compressor maps for every turbo depending on conditions). These factors should be insignificant as long as the pressure ratios do not vary widely, such as high boost (30psi) at extreme altitudes (10,000ft). When used within normal limits, this should help to make the correction factor on forced induction car a little more accurate.

MarkRobinson

ha, Clint's a good friend of mine & ex-customer: known him for about 9 years.

A smart cookie indeed, he still works with me/boostlogic & works for my tuning master: Justin @ tuning concepts. The combination of them assisted me with my turbokits. Great guys, all are ex-customers when I had my retail performance shop.