Thrust-to-weight ratio

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 = 1

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 celestial body without gliding or using parachutes, then its engines must put out more thrust than its current weight to counteract gravity. 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. 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.

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 without assistance from aerodynamic lift (i.e. wings). 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

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

Where:

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

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.

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

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

To estimate the maximum acceleration ($a$) at launching vertically only from knowing the TWR and gravitational acceleration the following formula can be used:

$a={\frac {F_{T}-F_{G}}{m}}={\frac {F_{T}-mg}{m}}={\frac {F_{T}}{m}}-g=g({\text{TWR}}-1)$

Where:

• ${\text{TWR}}$ the thrust-to-weight ratio for the given $g$
• The rest are the same from the original formula

Physical background

To lift off, the engines need to supply enough force in the opposite direction of the gravitational pull to counteract it. Usually the total thrust of all engines in the current stage running at full throttle is used in the calculation to find the largest possible ratio. 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 because of several reasons:

1. As the engines consume resources, the craft becomes lighter over time, raising the ratio.
2. On most engines the thrust can be throttled, so lowering the thrust leads to a lower ratio than one calculated for full throttle.
3. The gravitational pull is lower the farther from a body, so the ratio increases with altitude.
4. As previous stages are removed from a multistage craft, it becomes lighter as parts are removed and thrust changes as previous engines are removed and any subsequent engines start operating.
5. Docking and undocking will add or remove weight respectively, along with the possibility of adding or removing engines.
6. In an atmosphere the pressure changes with altitude and on most engines the specific impulse does too. When a craft ascends usually the specific impulse increases which increases since 1.0 the thrust. Before 1.0 the fuel flow decreased instead which does not influence the thrust and thus does not directly change the TWR but instead slows the increase of TWR (because less resources are consumed).

The engine is tilted by $\alpha =30^{\circ }$, 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:

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

Where:

• $F_{\mathit {eff}}$ is the effective thrust to counteract gravity
• $F_{T}$ is the engine's thrust
• $\alpha$ is the pitch of the engine (0° straight downward, 90° straight sideways)

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 130.94 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:

${\text{TWR}}={\frac {6\cdot 200{\text{kN}}+1500{\text{kN}}}{130.94{\text{t}}\cdot 9.81{\frac {\text{m}}{{\text{s}}^{2}}}}}=2.102$

A TWR of 2.102 is above 1 and means liftoff!

The second stage of a Kerbal X with a mass of 16.12 t and the Rockomax "Poodle" Liquid Engine with 220 kN thrust can lift off only with full throttle from Kerbin but it lifts off quite well from the Mun:

${\text{TWR}}_{\text{Kerbin}}={\frac {220{\text{kN}}}{16.12{\text{t}}\cdot g_{\text{Kerbin}}}}={\frac {220{\text{kN}}}{16.12{\text{t}}\cdot 9.81{\frac {\text{m}}{{\text{s}}^{2}}}}}=1.391$

${\text{TWR}}_{\text{Mun}}={\frac {220{\text{kN}}}{16.12{\text{t}}\cdot g_{\text{Mun}}}}={\frac {220{\text{kN}}}{16.12{\text{t}}\cdot 1.63{\frac {\text{m}}{{\text{s}}^{2}}}}}=8.373$

If the engine has been worked with the thrust of LV-909 Liquid Fuel Engine which produces only 50 kN thrust the stage itself wouldn't be able to lift off Kerbin, but still could lift up from Mun.

${\text{TWR}}_{\text{Kerbin}}={\frac {50{\text{kN}}}{16.12{\text{t}}\cdot g_{\text{Kerbin}}}}={\frac {50{\text{kN}}}{16.12{\text{t}}\cdot 9.81{\frac {\text{m}}{{\text{s}}^{2}}}}}=0.316$

${\text{TWR}}_{\text{Mun}}={\frac {50{\text{kN}}}{16.12{\text{t}}\cdot g_{\text{Mun}}}}={\frac {50{\text{kN}}}{16.12{\text{t}}\cdot 1.63{\frac {\text{m}}{{\text{s}}^{2}}}}}=1.903$

Practical illustration

Test craft massing 20 tonnes

The test craft shown here has a mass of 20 metric tons (20,000 kg); it is powered by a stock LV-T45 engine rated at 200 kN of thrust. As you can see, this yields a TWR at Kerbin surface just sufficient to lift off the pad.

Because the gravitational acceleration on Kerbin's surface is roughly 10 m/s², 10 kN per ton or 100 kg per unit of thrust result in a thrust-to-weight ratio of about 1. This represents the minimum for launch; a TWR in the range 1.5 to 2.5 is better.

See also

• Terminology
• Thrust-to-weight ratio on Wikipedia