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# Cheat sheet

 This page is in need of being brought up to date. Please help Kerbal Space Program Wiki by fixing inaccurate or outdated information. This page provides false information about the Δv required in planets with atmospheres. A lot of the information on this page has to be either removed or updated. Due to migration problems, The Δv map cannot be updated.

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.

## Mathematics

### Thrust-to-weight ratio (TWR)

This is Newton's Second Law. If the ratio is less than 1 the craft will not 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.

${\text{TWR}}={\frac {F_{T}}{m\cdot g}}>1$
Where:
• $F_{T}$ is the thrust of the engines
• $m$ the total mass of the craft
• $g$ the local gravitational acceleration (usually surface gravity)

### Combined specific impulse (Isp)

If the Isp is the same for all engines in a stage, then the Isp is equal to a single engine. If the Isp is different for engines in a single stage, then use the following equation:

$I_{{sp}}={\frac {(F_{1}+F_{2}+\dots )}{{\frac {F_{1}}{I_{{sp1}}}}+{\frac {F_{2}}{I_{{sp2}}}}+\dots }}$

### Delta-v (Δv)

#### Basic calculation

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

$\Delta {v}=ln\left({\frac {M_{{start}}}{M_{{end}}}}\right)\cdot I_{{sp}}\cdot 9.81{\frac {m}{s^{2}}}$
Where:
• $\Delta {v}$ is the velocity change possible in m/s
• $M_{{start}}$ is the starting mass in the same unit as $M_{{end}}$
• $M_{{end}}$ is the end mass in the same unit as $M_{{start}}$
• $I_{{sp}}$ is the specific impulse of the engine in seconds

#### True Δv of a stage that crosses from atmosphere to vacuum

Body Δvout
Kerbin 1000 m/s
other bodies' data missing

Calculation of a rocket stage's Δv, taking into account transitioning from atmosphere to vacuum. Δvout 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.

$\Delta {v}_{T}={\frac {\Delta {v}_{{atm}}-\Delta {v}_{{out}}}{\Delta {v}_{{atm}}}}\cdot \Delta {v}_{{vac}}+\Delta {v}_{{out}}$

#### 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.
ISP(Vac) (s) Max Δv (m/s) Engines
250 5394 O-10 "Puff"
290 6257 LV-1R "Spider"
24-77 "Twitch"
300 6473 KR-1x2 "Twin-Boar"
305 6581 CR-7 R.A.P.I.E.R.
Mk-55 "Thud"
310 6689 LV-T30 "Reliant"
RE-M3 "Mainsail"
315 6797 LV-1 "Ant"
KS-25 "Vector"
KS-25x4 "Mammoth"
320 6905 48-7S "Spark"
LV-T45 "Swivel"
RE-I5 "Skipper"
340 7336 KR-2L+ "Rhino"
T-1 "Dart"
345 7444 LV-909 "Terrier"
350 7552 RE-L10 "Poodle"
800 21837 LV-N "Nerv"
4200 33751 IX-6315 "Dawn"

(Version: 1.2.2)

## Math examples

### TWR

• Copy template:
TWR = F / (m * g) > 1

### Isp

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

$I_{{sp}}={\frac {(F_{1}+F_{2}+\dots )}{{\frac {F_{1}}{I_{{sp1}}}}+{\frac {F_{2}}{I_{{sp2}}}}+\dots }}$

• Simplified:
Isp = ( F1 + F2 + ... ) / ( ( F1 / Isp1 ) + ( F2 / Isp2 ) + ... )
• Explained:
Isp = ( Force of thrust of 1st engine + Force of thrust of 2nd engine...and so on... ) / ( ( Force of thrust of 1st engine / Isp of 1st engine ) + ( Force of thrust of 2nd engine / Isp of 2nd engine ) + ...and so on... )
• Example:
Two engines, one rated 200 newtons and 120 seconds Isp ; another engine rated 50 newtons and 200 seconds Isp.
Isp = (200 newtons + 50 newtons) / ( ( 200 newtons / 120 ) + ( 50 newtons / 200 ) = 130.4347826 seconds Isp

### Δv

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

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

• Simplified:
Δv = ln ( Mstart / Mdry ) * Isp * 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 Isp.
Δv = ln ( 23 Tons / 15 Tons ) × 120 seconds Isp × 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:
$\Delta {v}=21.576745349086\cdot I_{{sp}}$
• Simplified:
Δv =21.576745349086 * Isp
• 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 Isp.
Δv = ln ( 18 Tons / 2 Tons ) × 380 seconds Isp × 9.81m/s² = Maximum Δv of 8199.1632327878 m/s
Δv = 2.1972245773 × 380 seconds Isp × 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 Isp = 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 approximately 1000 m/s of Δv to escape Kerbin's atmosphere before vacuum Δv values take over for the stage powering the transition.
3. Note: This equation is an guess, 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:

$\Delta {v}_{T}={\frac {\Delta {v}_{{atm}}-\Delta {v}_{{out}}}{\Delta {v}_{{atm}}}}\cdot \Delta {v}_{{vac}}+\Delta {v}_{{out}}$

• Simplified:
True Δv = ( ( Δv atm - 1000 ) / Δv atm ) * Δv vac + 1000
• Explained:
True Δv = ( ( Total Δv in atmosphere - 1000 m/s) / Total Δv in atmosphere ) X Total Δv in vacuum + 1000
• Example:
Single stage with total atmospheric Δv of 5000 m/s, and rated 6000 Δv in vacuum.
Transitional Δv = ( ( 5000 Δv atm - 1000 Δv required to escape Kerbin atmosphere ) / 5000 Δv atm ) X 6000 Δv vac + 1000 Δv required to escape Kerbin atmosphere = Total Δv of 5800 m/s