The situation of a sphere falling in water is a classic fluid dynamics problem.  If the diameter and density of the sphere are known along with the fluid properties of water, then the terminal velocity of the sphere can be estimated.

Figure 1 - Sphere falling in water
Figure 1 shows the sphere at the water surface. The acceleration direction is shown in red.

Analytical Solution

If a creeping flow is assumed (Reynold's number < 1) then the terminal velocity is

U = (sphere_density - water_density)*sphere_diameter^2*g / (18*water_viscosity)

For an example let

Using the above values, the terminal velocity, assuming a creeping flow, is calculated as

U = (2,030 - 1,030)*(0.0001)^2*9.81 / (18*0.0020) = 0.002725 m/s

The calculated velocity must be checked with the Reynold's equation

Re = water_density*sphere_diameter*U / water_viscosity

Re = 1,030*0.0001*0.002725 / 0.002 = 0.140

Since Re < 1, the flow assumption is valid.

If the sphere diameter is larger, for example 0.1 m, then the velocity is calculated to be

U = (2,030 - 1,030)*(0.1)^2*9.81 / (18*0.0020)   = 0.002725 m/s


To set up the simulation, first define the simulation XML file.


<?xml version="1.0"?>
		<Engine engine="Newton" />
		<Gravity x="0" y="-9.81" z="0" />

	<Dimensions width="25" length="50" />
	<Texture file="../resources/bitmaps/water.bmp" />

		<Fog density="0.0" depth="100" />

	<WorldObject file="sphere.xml" />


<?xml version = "1.0"?> 
<WorldObject name = "sphere"> 
 <Origin x="0" y="0" z="0"/>
     <Model file="sphere.ms3d" /> 
     <Origin x="0" y="0" z="0" /> 
     <Scale x= "0.05" y= "0.05" z="0.05" />
    <Watch type="position" 
           interval="0.01" />
        <Sphere name="sphere">
            <Position x="0" y="0" z="0" />
            <Dimensions radius="0.0001" />
            <Mass mass= "0.0099999998" />


Figure One shows the position of the sphere plotting with respect to time.