A beam-energy-floor scan lowers beam energy across three rows toward a detection floor, showing how the delivered impulse falls with it row by row. The arithmetic is exact only for the stated energies, areas, times, and rows in the ideal uniform-beam model. Real measurements add beam profile, alignment, reflectivity/absorption, detector calibration, timing/area calibration, surface geometry, and uncertainty.

Finite beam time is scanned below, at, and above an energy 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

Finite beam energy is scanned through an energy floor

The irradiance source is 6 watts per square meter and illuminated area stays 2 square meters. The time window is 2 s. The asserted product gives 24 J which is below the 48 J floor.

E=IAt=6⋅2⋅2=24 JE=IAt=6\cdot2\cdot2=24\ \text{J}
Energy boundary rowThe finite time window moves the energy row.power=24 WbeamArea=4 m^2irradiance=6 W/m^2irradianceSource=boundilluminatedArea=2 m^2timeWindow=2 senergy=24 J

Finite beam energy is scanned through an energy floor

The irradiance source is 6 watts per square meter and illuminated area stays 2 square meters. The time window is 4 s. The asserted product gives 48 J which is at the 48 J floor.

E=IAt=6⋅2⋅4=48 JE=IAt=6\cdot2\cdot4=48\ \text{J}
Energy boundary rowThe finite time window moves the energy row.power=24 WbeamArea=4 m^2irradiance=6 W/m^2irradianceSource=boundilluminatedArea=2 m^2timeWindow=4 senergy=48 J

Finite beam energy is scanned through an energy floor

The irradiance source is 6 watts per square meter and illuminated area stays 2 square meters. The time window is 6 s. The asserted product gives 72 J which is above the 48 J floor.

E=IAt=6⋅2⋅6=72 JE=IAt=6\cdot2\cdot6=72\ \text{J}
Energy boundary rowThe finite time window moves the energy row.power=24 WbeamArea=4 m^2irradiance=6 W/m^2irradianceSource=boundilluminatedArea=2 m^2timeWindow=6 senergy=72 J