Understand the relationship
The reasoning behind the result
Why the mass ratio sits inside a logarithm
d(Δv) = −vₑ dm/m; Δv = vₑ ln(m₀/mf)
During a burn, propellant leaving the vehicle carries momentum away. A small expelled mass produces a velocity increment proportional to its fraction of the vehicle's current mass. That fraction changes as the vehicle becomes lighter, so multiplying one fixed acceleration by a time would miss the mass change.
Integrating the fractional mass change from initial mass m₀ to final mass mf produces the natural logarithm. Both masses must use the same unit: their quotient R = m₀/mf is dimensionless. Doubling both masses leaves ideal delta-v unchanged, although it doubles the propellant consumed.
Four quantities, four different meanings
vₑ = Isp × g₀; R = m₀/mf; mp = m₀ − mf; fp = 1 − 1/R
Isp is specific impulse in seconds; vₑ is effective exhaust velocity in metres per second. R describes the ratio of the whole vehicle's initial and final masses. Propellant mass mp is an absolute amount, while propellant fraction fp is a share of initial vehicle mass.
The result label always identifies the selected quantity. A velocity, a mass and a dimensionless ratio are not interchangeable outputs even when they are calculated from the same burn scenario.
Solving backward from a required delta-v
R = exp(Δv/vₑ); mp = mf × expm1(Δv/vₑ)
When the final mass is known, first determine the ratio needed for the entered velocity budget. Multiplying that ratio by final mass gives initial mass. Subtracting final mass gives expelled propellant. The implementation uses expm1 for the subtraction so very small budgets retain numerical precision.
When initial mass is fixed instead, final mass is m₀ exp(−Δv/vₑ), and the available mass left after the burn becomes a useful companion result. This answers a different starting-data question from fixing the delivered final mass.
Read the trade-off before adding propellant
With final mass and exhaust velocity held fixed, each additional unit of delta-v requires exponentially more initial mass. With a mass ratio held fixed, delta-v is proportional to effective exhaust velocity. These are mathematical sensitivities within the model; they do not account for changes in engine mass, tank mass or achievable performance.
The live mass balance shows what is expelled and what remains. It is drawn from the current valid inputs, while the supporting rows expose the mass ratio and exhaust velocity needed to check the calculation independently.
Follow the numbers
A 75% propellant fraction
- A fraction of 75% means fp = 0.75, leaving 1 − fp = 0.25 of initial vehicle mass.
- R = 1 / 0.25 = 4. If final mass is 1,000 kg, initial mass is 4,000 kg.
- Expelled propellant is 3,000 kg. With Isp = 300 s, ideal delta-v is 300 × 9.80665 × ln(4) = 4,078.471079 m/s.
A ratio of four means the initial vehicle is four times as massive as the remaining vehicle; it does not mean four times as much propellant as final mass.
Quick guide
How to use this calculator
- Select measured masses, a velocity budget or a propellant fraction.
- Use final mass for everything still aboard at burn completion.
- Read the dimensionless ratio alongside initial mass, propellant fraction and ideal delta-v.
Calculation method
Calculation and interpretation
Compare how much vehicle mass must be present before a burn relative to the mass that remains after it.
R = m₀/mf = exp(Δv/(Isp g₀)) = 1/(1 − fp)
Worked example
A 75% propellant fraction
A ratio of four means the initial vehicle is four times as massive as the remaining vehicle; it does not mean four times as much propellant as final mass.
R = m₀/mf = exp(Δv/(Isp g₀)) = 1/(1 − fp)
Supported inputs
Precision and limits
Ideal velocity budget
These are ideal, constant-effective-exhaust-velocity burns. Gravity loss, aerodynamic drag, steering loss, finite-burn orbital effects and engine transients are excluded. Delta-v is not an attained ground speed or a launch capability.
Mass definitions
Initial and final mass refer to the entire vehicle accelerated during one burn. Final mass includes structure, payload and any propellant that remains unburned. Expelled propellant is the difference between those masses.
Specific impulse convention
Specific impulse is entered in seconds. Its conversion to effective exhaust velocity uses standard gravity, 9.80665 m/s², even for a burn far from Earth. This constant is a unit convention, not a gravity-loss allowance.
Numerical and practical limits
A mass ratio must be at least one; final mass and exhaust velocity must be positive. A zero-propellant case has zero ideal delta-v. Extremely large required mass ratios may be numerically representable but physically impractical.
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