Illuminated area is scanned below, at, and above a force floor. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Illuminated area is scanned against a force floor

Pressure stays 1 Pa while illuminated area is 2 square meters. The asserted pressure-area product gives 2 N which is below the 3 N floor.

F=pA=12 m2=2 NF=pA=1\cdot2\ \text{m}^{2}=2\ \text{N}
Force boundary rowArea is the second checked input to force.power=24 WbeamArea=4 m^2irradiance=6 W/m^2irradianceSource=boundspeed=6 m/smode=absorberpressure=1 Padirection=normalpressureSource=boundilluminatedArea=2 m^2force=2 N

Illuminated area is scanned against a force floor

Pressure stays 1 Pa while illuminated area is 3 square meters. The asserted pressure-area product gives 3 N which is at the 3 N floor.

F=pA=13 m2=3 NF=pA=1\cdot3\ \text{m}^{2}=3\ \text{N}
Force boundary rowArea is the second checked input to force.power=24 WbeamArea=4 m^2irradiance=6 W/m^2irradianceSource=boundspeed=6 m/smode=absorberpressure=1 Padirection=normalpressureSource=boundilluminatedArea=3 m^2force=3 N

Illuminated area is scanned against a force floor

Pressure stays 1 Pa while illuminated area is 4 square meters. The asserted pressure-area product gives 4 N which is above the 3 N floor.

F=pA=14 m2=4 NF=pA=1\cdot4\ \text{m}^{2}=4\ \text{N}
Force boundary rowArea is the second checked input to force.power=24 WbeamArea=4 m^2irradiance=6 W/m^2irradianceSource=boundspeed=6 m/smode=absorberpressure=1 Padirection=normalpressureSource=boundilluminatedArea=4 m^2force=4 N