User:EsalazioeHg/Sandbox/Thrust-to-weight ratio/it La TWR è il rapporto tra FT e FG. F è rivolta verso l'alto se TWR > 1, verso il basso se TWR < 1 e nulla se TWR = 1

Il Rapporto spinta/peso (TWR, Thrust-to-Weight Ratio) è il rapporto che definisce la potenza del motore di una navicella in relazione al suo peso. Se una navicella deve raggiungere un'orbita stabile od atterrare in sicurezza sul corpo celeste attuale senza utilizzare paracadute, allora i suoi motori dovranno produrre una spinta maggiore del suo attuale peso per contrastare la gravità. In termini di rapporto, una navicella in grado di produrre una spinta maggiore del proprio peso avrà una TWR maggiore di 1. Il peso dipende dalla massa di questa e dall'accelerazione gravitazionale locale che solitamente è la gravità superficiale del corpo celeste nella cui sfera di influenza si trova l'astronave. In un'orbita stabile il rapporto spinta/peso non è importante, ma il suo valore può esser usato per stimare l'accelerazione massima possibile.

Se il rapporto è minore di 1 e la navicella è sulla superficie, allora questa non sarà in grado di decollare dal terreno senza l'ausilio di portanze aerodinamiche (i.e. profili alari). Viceversa, se questa sarà in caduta libera verso la superficie, i motori non potranno generare una spinta sufficiente ad ottenere un atterraggio morbido.

Formula

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

Quando la TWR e l'accelerazione gravitazionale di un corpo celeste (A) sono note, è possibile calcolare la TWR per la gravità superficiale di un altro corpo celeste (B). Nello specifico se la TWR conosciuta è relativa a Kerbin, è possibile utilizzare il valore della gravità superficiale espressa in g-force agente sul secondo corpo.

${\text{TWR}}_{A}\cdot {\frac {g_{B}}{g_{A}}}={\text{TWR}}_{B}$ ${\text{TWR}}_{\text{Kerbin}}\cdot g_{B}={\text{TWR}}_{B}$ , l'accelerazione gravitazionale $g_{B}$ è data in multipli di $g_{\text{Kerbin}}$ (g-force).

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:

$a={\frac {F_{T}-F_{G}}{m}}={\frac {F_{T}-mg}{m}}={\frac {F_{T}}{m}}-g=g({\text{TWR}}-1)$ Dove:
• ${\text{TWR}}$ 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 five reasons:

1. As the engines consume resources, the craft becomes lighter over time, raising the ratio
2. The gravitational pull is lower the farther from a body, so the ratio increases with 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
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. Both thrust and weight can increase from docking 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:

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

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.