Understand the relationship
The reasoning behind the result
Only the vertical thrust component balances weight
Fvertical=N Fengine cosθ; W=mg
The model assumes N identical active engines with equal thrust and a common angle θ from vertical. Only the vertical component contributes to this one-dimensional force balance. At zero cant, all axial thrust is vertical; at 60° only half is; at 90° the vertical component is zero.
The calculation does not model lateral force cancellation, torque, attitude control or vector differences among engines. Weight uses the entered local gravity and current vehicle mass. Standard gravity used in specific-impulse units must not be substituted automatically for that local value.
A thrust ratio is different from acceleration
TWR=Fvertical/W; avertical,ideal=(TWR−1)g
For positive local gravity, the ratio compares upward thrust with downward weight. In this ideal model a ratio below one gives a downward net acceleration, one gives zero vertical net force, and above one gives upward net acceleration.
Zero local gravity makes weight zero, so thrust-to-weight ratio is undefined rather than infinite performance. Thrust divided by mass still gives its acceleration component. The output preserves that distinction and does not replace an undefined ratio with an arbitrary large number.
A target ratio can have no finite engine solution
Fengine,target=TWRtarget·mg/(Ncosθ)
For positive gravity and at least one engine with a nonzero vertical component, solving the force balance gives the required thrust per identical active engine. Fewer active engines or more cant generally require greater per-engine axial thrust for the same vertical target.
No active engine cannot supply a positive thrust requirement, and fully horizontal engines cannot supply a positive vertical requirement under this model. These cases remain explicit statuses in the ledger. They do not silently become zero required thrust or a successful hover claim.
Follow the numbers
The effect of losing one active engine in an entered hover scenario
- For 1,000 kg under an entered local gravity of 1.62 m/s², weight is 1,620 N.
- Four identical vertical engines at a target ratio of one require 1,620/4=405 N per engine.
- With only three active engines, the same ideal vertical balance requires 1,620/3=540 N per engine. Both scenarios have zero ideal net vertical acceleration at exactly that thrust.
The arithmetic does not establish that either engine configuration has the control authority or operating margin to hover.
Quick guide
How to use this calculator
- Enter local gravitational acceleration; it is independent of the fixed g0 convention used for Isp.
- Record vehicle mass, active engine count and common engine cant angle for each configuration.
- Assess known per-engine thrust or specify a target vertical thrust/weight ratio.
- Read the ideal force balance and any undefined or impossible case. A ratio of one alone does not establish a controllable hover.
Calculation method
Calculation and interpretation
Separate total axial engine thrust, its vertical component, local vehicle weight and the resulting ideal vertical force balance.
W=mg; Fvertical=N·Fengine·cosθ; TWR=Fvertical/(mg); Fengine,target=TWRtarget·mg/(Ncosθ); avertical,ideal=Fvertical/m−g.
Worked example
The effect of losing one active engine in an entered hover scenario
The arithmetic does not establish that either engine configuration has the control authority or operating margin to hover.
W=mg; Fvertical=N·Fengine·cosθ; TWR=Fvertical/(mg); Fengine,target=TWRtarget·mg/(Ncosθ); avertical,ideal=Fvertical/m−g.
Supported inputs
Precision and limits
Instantaneous ideal vertical force balance
No drag, varying gravity, thrust transients, fuel depletion, vehicle rotation, lateral dynamics or control-system behavior is included. A positive net force or TWR above one is not a flight, landing or safety certification.
Identical active engines
All active engines in a row share the entered axial thrust and cant. This does not solve an unequal-thrust vector or torque distribution.
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