User:FancyGamer

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== What is Delta{v}

==

Definition \Delta{v} = \int_{t_0}^{t_1} {\frac {|T|} {m}}\, dt where

T is the instantaneous thrust m is the instantaneous mass Specific cases[edit] In the absence of external forces:

= \int_{t_0}^{t_1} {|a|}\, dt where a is the coordinate acceleration. When thrust is applied in a constant direction this simplifies to:

= | {v}_1 - {v}_0 |\;

which is simply the magnitude of the change in velocity. However, this relation does not hold in the general case: if, for instance, a constant, unidirectional acceleration is reversed after (t1 − t0)/2 then the velocity difference is 0, but delta-v is the same as for the non-reversed thrust.

For rockets the 'absence of external forces' is taken to mean the absence of gravity, atmospheric drag as well as the absence of aerostatic back pressure on the nozzle and hence the vacuum Isp is used for calculating the vehicle's delta-v capacity via the rocket equation, and the costs for the atmospheric losses are rolled into the delta-v budget when dealing with launches from a planetary surface.[citation needed]

Orbital maneuvers[edit] Main article: rocket equation Orbit maneuvers are made by firing a thruster to produce a reaction force acting on the spacecraft. The size of this force will be

T = V_{exh}\ \rho\,



(1) where

Vexh is the velocity of the exhaust gas ρ is the propellant flow rate to the combustion chamber The acceleration \dot{V}\, of the spacecraft caused by this force will be

\dot{V}=\frac{T}{m} = V_{exh}\ \frac{\rho}{m}\,



(2) where m is the mass of the spacecraft

During the burn the mass of the spacecraft will decrease due to use of fuel, the time derivative of the mass being

\dot{m}=-\rho\,



(3) If now the direction of the force, i.e. the direction of the nozzle, is fixed during the burn one gets the velocity increase from the thruster force of a burn starting at time t_0\, and ending at t1 as

\Delta{V} = -\int_{t_0}^{t_1} {V_{exh}\ \frac{\dot{m}}{m}}\, dt



(4) Changing the integration variable from time t to the spacecraft mass m one gets

\Delta{V} = -\int_{m_0}^{m_1} {V_{exh}\ \frac{dm}{m}}\,



(5) Assuming V_{exh}\, to be a constant not depending on the amount of fuel left this relation is integrated to

\Delta{V} = V_{exh}\ \ln(\frac{m_0}{m_1})\,



(6) which is the Tsiolkovsky rocket equation.

If for example 20% of the launch mass is fuel giving a constant V_{exh}\, of 2100 m/s (typical value for a hydrazine thruster) the capacity of the reaction control system is

\Delta{V} = 2100\ \ln(\frac{1}{0.8})\, m/s = 469 m/s. If V_{exh}\, is a non-constant function of the amount of fuel left[1]

V_{exh}=V_{exh}(m)\, the capacity of the reaction control system is computed by the integral (5)

The acceleration (2) caused by the thruster force is just an additional acceleration to be added to the other accelerations (force per unit mass) affecting the spacecraft and the orbit can easily be propagated with a numerical algorithm including also this thruster force.[2] But for many purposes, typically for studies or for maneuver optimization, they are approximated by impulsive maneuvers as illustrated in figure 1 with a \Delta{V}\, as given by (4). Like this one can for example use a "patched conics" approach modeling the maneuver as a shift from one Kepler orbit to another by an instantaneous change of the velocity vector.

Delta-v is typically provided by the thrust of a rocket engine, but can be created by other reaction engines. The time-rate of change of delta-v is the magnitude of the acceleration caused by the engines, i.e., the thrust per total vehicle mass. The actual acceleration vector would be found by adding thrust per mass on to the gravity vector and the vectors representing any other forces acting on the object.

The total delta-v needed is a good starting point for early design decisions since consideration of the added complexities are deferred to later times in the design process.

The rocket equation shows that the required amount of propellant dramatically increases, with increasing delta-v. Therefore in modern spacecraft propulsion systems considerable study is put into reducing the total delta-v needed for a given spaceflight, as well as designing spacecraft that are capable of producing a large delta-v.

--FancyGamer 04:29, 25 June 2015 (UTC)