Difference between revisions of "Thrust-to-weight ratio"

Revision as of 23:49, 22 August 2013

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

${\displaystyle {\text{TWR}}={\frac {F_{T}}{m\cdot g}}>1}$

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.

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

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

1. Because the engines consume fuel and the rocket get lighter meaning the ratio is rising over time
2. Because the gravity lowers with a higher distance to the celestial body the ratio is rising over time
3. Because the thrust can be throttled on certain engines the value can be modified in flight by the user the ratio may be lower than calculated

As soon as the rocket starts with the gravity turn only a portion of the craft's thrust is applied to counteract gravity, reducing the TWR. To calculate how much thrust is used to counteract gravity the pitch of the engine is important:

${\displaystyle F_{eff}=F_{T}\cdot \sin(\alpha )}$

This can also be used to calculate the thrust for already angled engines as they are already pitched.

Examples

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

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

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 and the LV-909 Liquid Fuel Engine with 50 kN thrust can not lift off from Kerbin but it can lift off from the Mun:

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

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

See also

• Terminology
• Thrust-to-weight ratio on Wikipedia