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

Revision as of 16:46, 8 October 2013

The TWR is the ratio of F_T and F_G. F is pointing upwards if the TWR > 1, downwards if TWR < 1 or doesn't exist if TWR = 0

The thrust-to-weight ratio (TWR) is a ratio that defines the power of a craft's engines in relation to its own weight. If a craft needs to get into a stable orbit or land safely on the current body without using parachutes, then its engines must put out more thrust than its current weight to counteract gravity. In a stable orbit, the thrust-to-weight ratio is not important, but it's value can be used to estimate the maximum acceleration possible. In the terms of a ratio, a craft with a greater thrust than weight will have a TWR greater than 1. The weight depends on the mass and local gravitational acceleration, which is usually the surface gravity of the body the craft is currently in the gravity well of.

If the ratio is less than 1 and the craft is on the surface, then the craft won't be able to lift off of the ground. If such a craft is currently falling towards the surface, then the craft's engines won't have enough thrust to slow down for a soft landing.

Formula

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

When the TWR and surface gravity for a celestial body (A) is known, it is possible to calculate the TWR for the surface gravity of another celestial body (B). Especially if the known TWR is for Kerbin, it is possible to use the surface gravity given in g-force acting on the second body.

${\displaystyle {\text{TWR}}_{A}\cdot {\frac {g_{B}}{g_{A}}}={\text{TWR}}_{B}}$

${\displaystyle {\text{TWR}}_{\text{Kerbin}}\cdot g_{B}={\text{TWR}}_{B}}$, the gravitational acceleration ${\displaystyle g_{B}}$ is given in multiples of ${\displaystyle g_{\text{Kerbin}}}$ (g-force).

To estimate the maximum acceleration (${\displaystyle a}$) only from knowing the TWR and gravitational acceleration the following formula can be used:

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

Physical background

To lift off, the engines need to supply enough force in the opposite direction of the gravitational pull to counteract it. The total thrust, or the force supplied by all 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 because of three reasons:

1. As the engines consume resources, the craft gets lighter over time, raising the ratio
2. The gravitational pull is lower the farther from a body, so the ratio is proportional to the altitude
3. On certain engines the thrust can be throttled, so lowering the thrust during flight leads to a lower ratio than one calculated for full throttle

The engine is tilted by ${\displaystyle \alpha =30^{\circ }}$, reducing the TWR

As soon as a craft 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 can be included:

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

This can also be used to calculate the thrust for engines that are placed angled on the craft. Technically it is like they are already pitched. Usually the engines on the other side are angled too, to thrust only upwards reducing the efficiency of the engines, because some thrust is cancelled out by them.

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:

${\displaystyle {\text{TWR}}={\frac {6\cdot 200{\text{kN}}+1500{\text{kN}}}{131.32{\text{t}}\cdot 9.81{\frac {\text{m}}{{\text{s}}^{2}}}}}=2.096}$

A TWR of 2.096 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:

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

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

See also

• Terminology
• Thrust-to-weight ratio on Wikipedia