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| − | [[File:Gravity turn start.svg|thumb|The TWR is the ratio of F<sub>T</sub> and F<sub>G</sub>. F is pointing upwards if the TWR > 1, downwards if TWR < 1 or doesn't exist if TWR = 1]]
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| − | 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 [[parachute]]s, 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 whose gravity well the craft is currently in. In a stable orbit, the thrust-to-weight ratio is not important, but its value can be used to estimate the maximum acceleration possible.
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| − | 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.
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| − | A useful equation to know is that your thrust has to be more than your mass multiplied by 9.81.
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| − | == Formula ==
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| − | {{Formula|math=\text{TWR} = \frac{F_T}{m \cdot g} > 1|where=* <math>F_T</math> is the thrust of the engines
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| − | * <math>m</math> the total mass of the craft
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| − | * <math>g</math> the local gravitational acceleration (usually surface gravity)}}
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| − | 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.
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| − | :<math>\text{TWR}_A \cdot \frac{g_B}{g_A} = \text{TWR}_B</math>
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| − | :<math>\text{TWR}_\text{Kerbin} \cdot g_B = \text{TWR}_B</math>, the gravitational acceleration <math>g_B</math> is given in multiples of <math>g_\text{Kerbin}</math> (g-force).
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| − | To estimate the maximum acceleration (<math>a</math>) at launching vertically only from knowing the TWR and gravitational acceleration the following formula can be used:
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| − | {{Formula|math=a = \frac{F_T-F_G}{m} = \frac{F_T-mg}{m} = \frac{F_T}{m} - g = g(\text{TWR}-1)|where=* <math>\text{TWR}</math> the thrust-to-weight ratio for the given <math>g</math>
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| − | * The rest are the same from the original formula}}
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| − | == Physical background ==
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| − | To lift off, the [[engine]]s 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.
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| − | :<math>\begin{align}
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| − | \sum\limits_i F_{T_\text{engine i}} = F_T &> F_G = m \cdot g \\
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| − | \frac{F_T}{m \cdot g} &> 1
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| − | \end{align}</math>
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| − | This value isn't constant over a flight because of several reasons:
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| − | # As the engines consume [[resource]]s, the craft becomes lighter over time, raising the ratio.
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| − | # On most engines the thrust can be throttled, so lowering the thrust leads to a lower ratio than one calculated for full throttle.
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| − | # The gravitational pull is lower the farther from a body, so the ratio increases with altitude.
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| − | # 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.
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| − | # Docking or undocking will add or remove weight respectively, along with the possibility of adding or removing engines.
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| − | # 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 the thrust since [[1.0]]. Before 1.0, the fuel flow decreased at low altitude instead which did not directly influence the thrust nor change the TWR but instead slowed the increase of TWR since less resources were consumed.
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| − | [[File:Gravity turn executed.svg|thumb|The engine is tilted by <math>\alpha = 30^\circ</math>, reducing the TWR]]
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| − | 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:
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| − | {{Formula|math=F_\mathit{eff} = F_T \cdot \cos(\alpha)|where=* <math>F_\mathit{eff}</math> is the effective thrust to counteract gravity
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| − | * <math>F_T</math> is the engine's thrust
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| − | * <math>\alpha</math> is the pitch of the engine (0° straight downward, 90° straight sideways)}}
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| − | 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.
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| − | == Examples ==
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| − | The [[Kerbal X]] with a mass of 130.94 t, 6 [[LV-T45 Liquid Fuel Engine]]s and 1 [[Rockomax "Mainsail" Liquid Engine]] on the [[launch pad]] of the [[Kerbal Space Center]] has a TWR of:
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| − | :<math>\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</math>
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| − | A TWR of 2.102 is above 1 and means liftoff!
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| − | 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]]:
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| − | :<math>\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</math>
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| − | :<math>\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</math>
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| − | 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.
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| − | :<math>\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</math>
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| − | :<math>\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</math>
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| − | == Practical illustration ==
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| − | [[File:TWR-test-Whose.png|288px|thumb|Test craft massing 20 tonnes]]
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| − | 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.
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| − | 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.
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| − | == See also ==
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| − | * [[Terminology#TWR|Terminology]]
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| − | * {{Wikipedia}}
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| − | [[Category:Physics]]
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