A radiation-impulse-floor scan lowers force or exposure time across three rows, checking when J = F * t drops below the stated impulse floor. The arithmetic is exact only for the stated forces, times, impulse floor, and rows in the ideal uniform-beam interaction model. Real measurements add beam profile, alignment, reflectivity/absorption, detector calibration, timing/area calibration, surface geometry, finite precision, and uncertainty.

Force is scanned below, at, and above a finite impulse 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

Force is scanned through a finite impulse floor

The time window stays 4 s while force is 2 N. The asserted impulse is 8 kilogram meters per second which is below the 12 kilogram meters per second impulse floor.

J=Ft=2⋅4=8 kg m/sJ=Ft=2\cdot4=8\ \text{kg}\,\text{m}/\text{s}
Impulse boundary rowForce changes while the finite time window stays fixed.power=24 WbeamArea=4 m^2irradiance=6 W/m^2irradianceSource=boundspeed=6 m/smode=absorberpressure=1 Padirection=normalpressureSource=boundilluminatedArea=2 m^2force=2 NforceSource=boundtimeWindow=4 simpulse=8 kg*m/smomentum=0 kg*m/sshowMomentumBit=0 bit

Force is scanned through a finite impulse floor

The time window stays 4 s while force is 3 N. The asserted impulse is 12 kilogram meters per second which is at the 12 kilogram meters per second impulse floor.

J=Ft=3⋅4=12 kg m/sJ=Ft=3\cdot4=12\ \text{kg}\,\text{m}/\text{s}
Impulse boundary rowForce changes while the finite time window stays fixed.power=24 WbeamArea=4 m^2irradiance=6 W/m^2irradianceSource=boundspeed=6 m/smode=absorberpressure=1 Padirection=normalpressureSource=boundilluminatedArea=3 m^2force=3 NforceSource=boundtimeWindow=4 simpulse=12 kg*m/smomentum=0 kg*m/sshowMomentumBit=0 bit

Force is scanned through a finite impulse floor

The time window stays 4 s while force is 4 N. The asserted impulse is 16 kilogram meters per second which is above the 12 kilogram meters per second impulse floor.

J=Ft=4⋅4=16 kg m/sJ=Ft=4\cdot4=16\ \text{kg}\,\text{m}/\text{s}
Impulse boundary rowForce changes while the finite time window stays fixed.power=24 WbeamArea=4 m^2irradiance=6 W/m^2irradianceSource=boundspeed=6 m/smode=absorberpressure=1 Padirection=normalpressureSource=boundilluminatedArea=4 m^2force=4 NforceSource=boundtimeWindow=4 simpulse=16 kg*m/smomentum=0 kg*m/sshowMomentumBit=0 bit