Cheat sheet
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.
Contents
Mathematics
Thrust-to-weight ratio (TWR)
- → See also: Thrust-to-weight ratio
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.
- is the thrust of the engines
- the total mass of the craft
- the local gravitational acceleration (usually surface gravity)
Combined specific impulse (I_{sp})
- → See also: Specific impulse
If the I_{sp} is the same for all engines in a stage, then the I_{sp} is equal to a single engine. If the I_{sp} is different for engines in a single stage, then use the following equation:
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.
- is the velocity change possible in m/s
- is the starting mass in the same unit as
- is the end mass in the same unit as
- is the specific impulse of the engine in seconds
True Δv of a stage that crosses from atmosphere to vacuum
Body | Δv_{out} |
---|---|
Kerbin | 2500 m/s |
other bodies' data missing |
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.
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 17261 LV-N "Nerv" 4200 33751 IX-6315 "Dawn"
(Version: 1.2.2)
Math examples
TWR
- Copy template:
- TWR = F / (m * g) > 1
I_{sp}
- 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}.
- When I_{sp} is different for engines in a single stage, then use the following equation:
- Equation:
- Simplified:
- I_{sp} = ( F1 + F2 + ... ) / ( ( F1 / I_{sp}1 ) + ( F2 / I_{sp}2 ) + ... )
- 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
- For atmospheric Δv value, use atmospheric values.
- For vacuum Δv value, use vacuum values.
- Use this equation to figure out the Δv per stage:
- Equation:
- 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:
- 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
- How to calculate the Δv of a rocket stage that transitions from Kerbin atmosphere to vacuum.
- 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.
- 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:
- 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