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Kerbal Space Program rocket scientist's cheat sheet: Delta-v maps, equations and more for your reference so you can get from here to there and back again.

• 1Mathematics
  • 1.3Delta-v (Δv)
• 2Math examples

Mathematics

Thrust-to-weight ratio (TWR)

→ See also: Thrust-to-weight ratio

How hard do your engines push UP? That's 'thrust'.

How hard does gravity pull DOWN? That's 'weight'.

The ratio of those two is, surprisingly, the 'thrust-to-weight' ratio.

If thrust is pushing harder then weight, your rocket goes up! Hooray!

If weight is pulling harder than thrust, your rocket does not go up. Sad!

That's the 'thrust-to-weight' ratio, or TWR, in a nutshell.

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

If TWR is less than one, then your rocket will not be going into space. You will need more engines, or perhaps MOAR BOOSTERZ!

If TWR is significantly above, say, 2.0, your rocket will ZOOM! But it might ZOOM too hard for your poor Kerbalnauts. Most rockets, straight off the launch pad, will utilize a TWR between 1.5 and 2.0.

Specific Impulse (I_sp)

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→ See also: Specific impulse

Specific impulse measures the efficiency of a particular engine. Because this depends on the pressure of a rocket's surroundings, this can change between Sea Level and Vacuum (in space). Most engines are designed for optimal performance either at Sea Level (the big ones that drive your first stage and boosters, which have lots of thrust!) or Vacuum (where they won't burn as much propellant, but can be VERY efficient).

Specific impulse is used in important equations like THE ROCKET EQUATION (see next section) to determine how much oomph (or 'delta-vee') you can get out of a given amount of propellant. The higher the specific impulse, the more efficient an engine is!

In real life, specific impulse is governed by things like combustion chamber pressure and propellant energy (hyrdogen/oxygen is more efficient than kerosene/oxgyen). Fortunately for Jebediah, our dear Kerbals only have one brand of fuel to worry about.

Delta-v (Δv)

Basic calculation

→ See also: Tutorial:Advanced Rocket Design

Basic calculation of a rocket's Δv. Use the atmospheric and vacuum thrust values for atmospheric and vacuum Δv, respectively.

• ${\displaystyle \mathrm{\Delta }v}$ is the velocity change possible in m/s
• ${\displaystyle M_{start}}$ is the starting mass in the same unit as ${\displaystyle M_{end}}$
• ${\displaystyle M_{end}}$ is the end mass in the same unit as ${\displaystyle M_{start}}$
• ${\displaystyle I_{sp}}$ is the specific impulse of the engine in seconds

True Δv of a stage that crosses from atmosphere to vacuum

Calculation of a rocket stage's Δv, taking into account transitioning from atmosphere to vacuum. Δv_out is the amount of Δv required to leave a body's atmosphere, not reach orbit. This equation is useful to figure out the actual Δv of a stage that transitions from atmosphere to vacuum.

${\displaystyle \mathrm{\Delta }{v}_T=\frac{\mathrm{\Delta }{v}_{atm}-\mathrm{\Delta }{v}_{out}}{\mathrm{\Delta }{v}_{atm}}\cdot \mathrm{\Delta }{v}_{vac}+\mathrm{\Delta }{v}_{out}}$

Maps

Various fan-made maps showing the Δv required to travel to a certain body.

Subway style Δv map (KSP 1.2.1):

Total Δv values

Δv change values

Δv with Phase Angles

Precise Total Δv values

WAC's Δv Map for KSP 1.0.4

Maximum Δv chart

This chart is a quick guide to what engine to use for a single stage interplanetary ship. No matter how much fuel you add you will never reach these ΔV without staging to shed mass or using the slingshot maneuver. (These calculations use a full/empty fuel-tank mass ratio of 9 for all engines except those noted.)

(Version: 1.6.1)

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Math examples

TWR

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• Copy template:

TWR = F / (m * g) > 1

I_sp

1. When I_sp is the same for all engines in a stage, then the I_sp is equal to a single engine. So six 200 I_sp engines still yields only 200 I_sp.
2. When I_sp is different for engines in a single stage, then use the following equation:

• Equation:

${\displaystyle I_{sp}=\frac{(F_1+F_2+\dots )}{\frac{F_1}{I_{sp1}}+\frac{F_2}{I_{sp2}}+\dots }}$

• Simplified:

I_sp = ( F1 + F2 + ... ) / ( ( F1 / I_sp1 ) + ( F2 / I_sp2 ) + ... )

• Explained:

I_sp = ( Force of thrust of 1st engine + Force of thrust of 2nd engine...and so on... ) / ( ( Force of thrust of 1st engine / I_sp of 1st engine ) + ( Force of thrust of 2nd engine / I_sp of 2nd engine ) + ...and so on... )

• Example:

Two engines, one rated 200 newtons and 120 seconds I_sp ; another engine rated 50 newtons and 200 seconds I_sp.

Isp = (200 newtons + 50 newtons) / ( ( 200 newtons / 120 ) + ( 50 newtons / 200 ) = 130.4347826 seconds I_sp

Δv

1. For atmospheric Δv value, use atmospheric ${\displaystyle I_{sp}}$ values.
2. For vacuum Δv value, use vacuum ${\displaystyle I_{sp}}$ values.
3. Use this equation to figure out the Δv per stage:

• Equation:

${\displaystyle \mathrm{\Delta }v=ln\left(\frac{M_{start}}{M_{dry}}\right)\cdot I_{sp}\cdot 9.81\frac{m}{s^2}}$

• Simplified:

Δv = ln ( Mstart / Mdry ) * I_sp * g

• Explained:

Δv = ln ( starting mass / dry mass ) X Isp X 9.81

• Example:

Single stage rocket that weighs 23 tons when full, 15 tons when fuel is emptied, and engine that outputs 120 seconds I_sp.

Δv = ln ( 23 Tons / 15 Tons ) × 120 seconds I_sp × 9.81m/s² = Total Δv of 503.0152618 m/s

Maximum Δv

Simplified version of the Δv calculation to find the maximum Δv a craft with the given ISP could hope to achieve. This is done by using a magic 0 mass engine and not having a payload.

• Equation:

${\displaystyle \mathrm{\Delta }v=21.576745349086\cdot I_{sp}}$

• Simplified:

Δv =21.576745349086 * I_sp

• Explained / Examples:

This calculation only uses the mass of the fuel tanks and so the ln ( Mstart / Mdry ) part of the Δv equation has been replaced by a constant as Mstart / Mdry is always 9 (or worse with some fuel tanks) regardless of how many fuel tanks you use.

The following example will use a single stage and fuel tanks in the T-100 to Jumbo 64 range with an engine that outputs 380 seconds I_sp.

Δv = ln ( 18 Tons / 2 Tons ) × 380 seconds I_sp × 9.81m/s² = Maximum Δv of 8199.1632327878 m/s

Δv = 2.1972245773 × 380 seconds I_sp × 9.82m/s² = Maximum Δv of 8199.1632327878 m/s (Replaced the log of mass with a constant as the ratio of total mass to dry mass is constant regardless of the number of tanks used as there is no other mass involved)

Δv = 21.576745349086 × 380 seconds I_sp = Maximum Δv of 8199.1632327878 m/s (Reduced to its most simple form by combining all the constants)

True Δv

1. How to calculate the Δv of a rocket stage that transitions from Kerbin atmosphere to vacuum.
2. Assumption: It takes roughly 2500 m/s of Δv to escape Kerbin's atmosphere before vacuum Δv values take over for the stage powering the transition (actual value ranges between 2000 m/s and 3400 m/s depending on ascent). Note that, as of KSP 1.3.1, around 3800 m/s of Δv is required to reach an 80km orbit from the KSC.
3. Note: This equation is a guess, an approximation, and is not 100% accurate. Per forum user stupid_chris who came up with the equation: 'The results will vary a bit depending on your TWR and such, but it should usually be pretty darn accurate.'

• Equation for Kerbin atmospheric escape:

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${\displaystyle \mathrm{\Delta }{v}_T=\frac{\mathrm{\Delta }{v}_{atm}-\mathrm{\Delta }{v}_{out}}{\mathrm{\Delta }{v}_{atm}}\cdot \mathrm{\Delta }{v}_{vac}+\mathrm{\Delta }{v}_{out}}$

• Simplified:

True Δv = ( ( Δv atm - 2500 ) / Δv atm ) * Δv vac + 2500

• Explained:

True Δv = ( ( Total Δv in atmosphere - 2500 m/s) / Total Δv in atmosphere ) X Total Δv in vacuum + 2500

• Example:

Single stage with total atmospheric Δv of 5000 m/s, and rated 6000 Δv in vacuum.

Transitional Δv = ( ( 5000 Δv atm - 2500 Δv required to escape Kerbin atmosphere ) / 5000 Δv atm ) X 6000 Δv vac + 2500 Δv required to escape Kerbin atmosphere = Total Δv of 5500 m/s

See also