Boost does not make power — density does
An engine makes power in proportion to the mass of air it traps per cycle, because the fuel it can burn is set by that air mass. A compressor raises pressure, which raises density, which raises trapped mass. But compressing air also heats it, and hot air is less dense, so part of every pressure gain is immediately given back as a temperature loss.
The ideal gas law makes the trade explicit: density is proportional to pressure divided by absolute temperature. Doubling the absolute pressure doubles the density only if the temperature does not change. Compress air to a pressure ratio of 2.0 through a 70% efficient compressor and the absolute temperature rises by about 31%, so the density ratio comes out at 1.52 rather than 2.00 — almost half the potential gain lost to heat.
That is why an intercooler is not an accessory. Removing 70% of the temperature rise recovers most of the lost density, and the same 2.00 pressure ratio climbs back to 1.83. Charge cooling also buys detonation margin, which usually lets you run more ignition advance and a leaner mixture, so the real-world gain exceeds the density arithmetic.
Gauge boost on its own is therefore a poor specification. Ten psi at sea level is a pressure ratio of 1.68; ten psi at 8,000 feet is a ratio of 1.92, because the denominator shrank. Always work in absolute pressure ratio, which is also what you need for a compressor map.
Isentropic compression and what efficiency actually means
Compressing a gas without heat transfer follows the isentropic relation T2s/T1 = PR(k−1)/k, where k is the ratio of specific heats — 1.4 for air, making the exponent 0.2857. That gives the temperature rise a perfect compressor would produce.
No compressor is perfect. Isentropic efficiency ηc is the ratio of the ideal temperature rise to the actual one, so the real discharge temperature is:
T2 = T1 × [1 + (PR0.2857 − 1) / ηc]
A 70% efficient compressor produces a temperature rise 1/0.70 = 1.43 times the ideal one. The extra heat is wasted work, and it is why efficiency islands on a compressor map matter so much: running at 60% instead of 74% at the same pressure ratio can cost 40°F of charge temperature and several percent of density.
Roots-type superchargers are the extreme case. A classic Roots blower does no internal compression at all — it displaces air into an already pressurised manifold — so its adiabatic efficiency at road-going pressure ratios is often in the 50 to 60% band, and discharge temperatures are correspondingly high. Screw superchargers and centrifugals do compress internally and run considerably better.
Intercooler effectiveness is defined as the fraction of the available temperature difference the core removes: ε = (T2 − T3) / (T2 − Tcoolant). This calculator uses ambient air as the coolant, which is the air-to-air case. A 70% effective core at 180°F inlet and 80°F ambient delivers 180 − 0.70 × 100 = 110°F. Effectiveness above about 80% is difficult in an air-to-air core at road speeds; air-to-water systems with ice can briefly exceed it and can even cool below ambient, which this model does not represent.
Worked example: 350 hp engine at 8 psi
A 350 hp naturally aspirated engine at sea level, 80°F ambient, fitted with a turbo running 8 psi through a 70% efficient compressor and no intercooler.
- Absolute manifold pressure. 8 + 14.696 = 22.696 psia.
- Pressure ratio. 22.696 ÷ 14.696 = 1.5444.
- Ideal temperature ratio. 1.54440.2857 = 1.1322, a 13.22% rise in absolute temperature if the compressor were perfect.
- Actual temperature ratio. 1 + 0.1322 ÷ 0.70 = 1.1889.
- Discharge temperature. Inlet is 80 + 459.67 = 539.67 °R, so T2 = 539.67 × 1.1889 = 641.6 °R = 181.9°F.
- Density ratio. 1.5444 × (539.67 ÷ 641.6) = 1.5444 × 0.8411 = 1.2990.
- Power. 350 × 1.2990 = 454.7 hp.
Notice the shortfall: the pressure ratio was 1.544 but the density ratio only 1.299. The 8 psi should have been worth 190 hp and delivered 105. The missing 85 hp went up the intake pipe as heat.
Now add a 70% effective intercooler. The charge falls to 181.9 − 0.70 × (181.9 − 80) = 110.6°F, which is 570.3 °R. The density ratio becomes 1.5444 × (539.67 ÷ 570.3) = 1.4614, and the power estimate rises to 511.5 hp. The intercooler was worth 57 hp at 8 psi, and it costs nothing in fuel.
How to read the estimate and where it breaks
This is a density model, so it answers one question well: how much more air is in the cylinder. Treat the horsepower number as an upper bound that a well-tuned engine approaches, not a promise.
Three effects push the real result below it. Exhaust backpressure from a turbine raises pumping work and dilutes the fresh charge with residual gas; on a poorly matched turbine, manifold backpressure can exceed boost pressure and the engine loses more than the compressor gains. Ignition retard is applied by every knock-controlled ECU as charge temperature and cylinder pressure rise, and retarded timing costs torque directly. Enrichment under boost cools the charge chemically but moves the mixture away from best-power stoichiometry. The derate input exists to let you apply a realistic haircut to the gain — 90% is a reasonable starting point for a street setup on pump fuel.
One effect pushes the other way: a supercharger or turbocharger changes the effective volumetric efficiency, and a boosted engine scavenges differently at overlap. Turbo engines with good scavenging can trap more air than the pressure ratio alone implies.
Charge temperature is the number to watch. Above roughly 160°F entering the cylinder, most pump-fuel engines start losing timing to knock control, and above 200°F the loss is severe. If your result shows a high charge temperature, more intercooler capacity will usually gain more real power than more boost. That is also the point at which alcohol fuel or water-methanol injection starts to pay, since both cool the charge as they vaporise — and both change your injector sizing.
Finally, verify that the compression ratio suits the boost. High static compression plus high boost is how head gaskets leave. Run the numbers with the compression ratio calculator before you order pistons.
Density ratio at sea level, 80°F inlet, 70% compressor efficiency
| Boost (psi) | Pressure ratio | Discharge temp (°F) | Density ratio, no intercooler | Density ratio, 70% intercooler |
|---|---|---|---|---|
| 5 | 1.340 | 147.3 | 1.192 | 1.292 |
| 8 | 1.544 | 181.9 | 1.299 | 1.462 |
| 10 | 1.680 | 203.2 | 1.368 | 1.573 |
| 15 | 2.021 | 251.7 | 1.533 | 1.845 |
| 20 | 2.361 | 294.5 | 1.690 | 2.110 |
| 25 | 2.701 | 333.1 | 1.839 | 2.368 |
| 30 | 3.042 | 368.4 | 1.982 | 2.621 |
Computed with k = 1.4 and an inlet of 80°F at 14.696 psia. Real compressors lose efficiency at the extremes of their maps, so the high-boost rows assume more than most single turbochargers deliver.
Assumptions and limits of this model
- It scales a measured baseline. If your naturally aspirated figure is wrong, everything downstream is wrong by the same proportion.
- It ignores exhaust backpressure. A turbine that is too small can cost more in pumping work than the compressor gains in density. Use the derate input to represent it.
- Compressor efficiency is not a constant. It varies across the map with flow and pressure ratio. The single number you enter is only valid at one operating point.
- Intercooler effectiveness falls with airflow. A core that is 75% effective at 60 mph may be far less on a dyno or in traffic, and pressure drop across it reduces manifold pressure.
- It assumes fuel and octane keep up. The density model does not know whether the fuel system can supply the mass or the fuel can resist knock at that pressure.
- It ignores charge cooling from fuel evaporation. Port-injected alcohol fuels cool the charge significantly after the intercooler, so real density in the cylinder can exceed this estimate.
Matching the hardware to the number
Once you have a target density ratio, the next question is whether a given compressor can supply it. That takes two coordinates: pressure ratio, and mass flow in pounds per minute. Pressure ratio comes from the pressure ratio calculator, which handles inlet restriction and intercooler pressure drop properly. Mass flow follows from displacement, rpm and volumetric efficiency — the same airflow arithmetic behind induction sizing, converted from volume to mass. Plot the pair on the compressor map and check that it lands inside a high-efficiency island and clear of the surge line.
Superchargers and turbochargers reach the same density by different routes. A positive-displacement supercharger delivers boost from just off idle and is driven by the crank, so it costs perhaps 15 to 20% of its output in parasitic drive power; that loss is already inside your dyno-measured result if you measure the blown engine, but it must be accounted for if you are predicting. A turbocharger takes its energy from exhaust enthalpy that would otherwise be wasted, at the cost of backpressure and lag. A centrifugal supercharger behaves like a turbo compressor driven by a belt, so its boost rises with the square of engine speed.
Whichever route you take, the supporting systems scale with the density ratio, not the pressure ratio. If the density ratio is 1.46, the engine is swallowing 46% more air, so it needs 46% more fuel — and the injectors and pump must both cover it. The connecting rods and head fasteners see a similar rise in peak cylinder pressure. Compression ratio, cam timing and fuel octane then decide how much of that pressure the engine will tolerate before knock control takes the timing away. Verify what you actually built at the strip using the trap speed method — it is the cheapest independent check on any boosted power claim.