Difference between revisions of "Thrust-to-weight ratio"
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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]]: | 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]]: | ||
− | :<math>\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</math> | + | :<math>\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</math> |
− | :<math>\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</math> | + | :<math>\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</math> |
== See also == | == See also == |
Revision as of 14:24, 28 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
- is the thrust of the engines
- the total mass of the craft
- 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.
This value isn't constant over a flight for three reasons:
- Because the engines consume fuel and the rocket get lighter, the ratio is rising over time
- Because the gravity lowers with a higher altitude, the ratio is proportional to the altitude
- Because on certain engines the thrust can be throttled, modified thrust in flight lead to a lower ratio 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 can be included:
- is the effective thrust to counteract gravity
- is the engine's thrust
- is the pitch of the engine (0° = straight forward, 90° straight downward)
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:
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:
See also
- Terminology
- Thrust-to-weight ratio on Wikipedia