How this calculation works
Subtract the weight of displaced fluid from object weight, then equate the remaining downward force to quadratic drag. Solve for a positive speed magnitude; the object must be denser than its displaced fluid.
vterminal = √[2(m−ρV)g/(ρCdA)]A worked example
A 2 kg object displacing 1 L in a 1,000 kg/m³ fluid has net downward force 10 N at g = 10. With Cd = 1 and reference area 0.01 m², the modeled terminal speed is √2 m/s.
Good to know
Fully immersed constant-property downward quadratic-drag model. A rising or neutrally buoyant object is outside this scope. No transient motion, speed-dependent Cd, Stokes law or safety certification is supplied.
Reference: NASA Glenn: terminal speed from force balance (opens in a new tab) ↗
Balance three forces at steady descent
A fully immersed object feels downward weight, upward buoyancy and upward drag while settling. Terminal speed occurs when those forces balance, leaving zero vertical acceleration under the stated constant-property model.
In the example, object weight is 20 N and displaced-fluid weight is 10 N. Drag therefore needs to supply only the remaining 10 N. With ρCdA/2 = 5 in consistent SI units, v² = 10/5 = 2 and speed is √2 m/s.
The buoyancy and displacement calculator separates displaced-fluid weight before this page uses it in a downward force balance.
Displacement changes the available downward force
The buoyancy and displacement calculator examines the displaced-fluid weight separately. Here it is subtracted before solving the drag balance. Neglecting buoyancy can matter greatly when object and fluid densities are similar.
Volume zero is an explicit way to neglect buoyancy in the entered scenario. A nonzero volume should represent the fully immersed displaced volume, not the container volume or the object’s drag reference area.
A force balance is not a fall-time simulation
The table evaluates drag and the remaining net force at fractions of terminal speed. It does not integrate the motion or report a distance needed to approach the terminal value. Constant Cd and orientation are assumptions throughout.
If object mass does not exceed displaced-fluid mass, this downward-settling model is inapplicable. A neutrally buoyant or rising object has a different direction question. The quadratic drag force calculator can evaluate a specified relative speed without implying a settling trajectory.
A few questions, answered
Why reject an object lighter than its displaced fluid?
It has no positive downward force for this settling balance; an upward-motion model would be needed.
Is this Stokes terminal velocity?
No. It uses a supplied constant quadratic drag coefficient, not the linear viscous-drag law for very low-Reynolds-number spheres.
Does terminal mean the maximum safe speed?
No. It describes a model force balance and supplies no safety rating.