Thrust-to-weight ratio

The thrust-to-weight ratio defines if a rocket can lift off the ground. If the ratio is less than 1 the craft can't lift off the ground. Note that the local gravitational acceleration, which is usually the surface gravity of the body the rocket is starting from, is required.

Formula

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Where:

• $F_{T}$ is the thrust of the engines
• $m$ the total mass of the craft
• $g$ the local gravitational acceleration (usually surface gravity)

Physical background

To lift off, the engines need to supply enough force to counteract the gravitational pull. The thrust, meaning the force supplied by the engines, is the sum of the thrust of all running engines. Usually the maximum thrust is used to know the upper limits. The gravitational pull is the weight of the craft which can be calculated by multiplying the mass with the current gravitation. To make the formula easier the surface gravity of the celestial body in question is used.

${\begin{aligned}\sum \limits _{i}F_{T_{\text{engine i}}}=F_{T}&>F_{G}=m\cdot g\\{\frac {F_{T}}{m\cdot g}}&>1\end{aligned}}$

This value isn't constant over a flight for three reasons:

1. Because the engines use fuel the rocket get lighter meaning the ratio is rising over time
2. Because the gravity changes on the distance to the celestial body's sphere of influence the craft is in
3. Because the thrust can be throttled on certain engines the value can be modified in flight by the user

Examples

The Kerbal X with a mass of 131.32 t, 6 LV-T45 Liquid Fuel Engines and 1 Rockomax "Mainsail" Liquid Engine has on the launch pad of the Kerbal Space Center a TWR of:

${\text{TWR}}={\frac {6\cdot 200{\text{kN}}+1500{\text{kN}}}{131.32{\text{t}}\cdot 9.81{\frac {\text{m}}{{\text{s}}^{2}}}}}=2.872$

A TWR of 2.872 is above 1 and means liftoff!

The third stage of a Kerbal X with a mass of 16.52 t but the LV-909 Liquid Fuel Engine with 50 kN thrust instead can't lift of Kerbin but of the Mun:

${\text{TWR}}={\frac {50{\text{kN}}}{16.52{\text{t}}\cdot g_{\text{Kerbin}}}}={\frac {50{\text{kN}}}{16.52{\text{t}}\cdot 9.81{\frac {\text{m}}{{\text{s}}^{2}}}}}=0.309$

${\text{TWR}}={\frac {50{\text{kN}}}{16.52{\text{t}}\cdot g_{\text{Mun}}}}={\frac {50{\text{kN}}}{16.52{\text{t}}\cdot 1.63{\frac {\text{m}}{{\text{s}}^{2}}}}}=1.856$

See also

• Terminology
• Thrust-to-weight ratio on Wikipedia