# Determine velocity of balloon from known Buoyancy, mass and surface [closed]

I need to know how to get velocity of weather balloon, if I know it's buoyancy force, mass and surface(diameter). I am using now $$v = 0.5t\left(\frac{F_\text{all}}{m}\right),$$ [$m$ = mass, $v$ = velocity, $t$ = overall time in seconds], when
$$F_\text{all} = F_\text{buoyancy} - mg - F_\text{air resistance}.$$ But $F_{air\ resistance}$ is too large (sometimes $7000\ \rm N$), it's larger then $F$ but it is without logic to $F$ be smaller then air resistance. For air resistance I am using equation from wikipedia.

I am using it in C# program and the velocity is normal from start, but then it goes to deep negative values. Can somebody give me working equation or say me where are my errors?

C# code is:

//t = overall time in seconds, F = overall force and hmotnost is mass
v = 0.5 * t * (F / hmotnost);
//Ov is drag(air resistance) hustotaV is air density, r is radius
Ov = 0.5 * hustotaV * Math.Pow(v, 2) * 0.5 * (4 * Math.PI * Math.Pow(r, 2) / 2);
//Fvz is uplift(buoyancy)
Fvz = ((4.0 / 3) * Math.PI * Math.Pow(r, 3)) * hustotaV * 9.80665;
//Ft is mass * 9.81
F = Fvz - Ft - Ov;

• defj, in your starting equation, Fall/m is an acceleration term. The balloon will quickly reach a "terminal" velocity when the air drag plus weight equals the buoyant force. Due to this, acceleration will be zero, and velocity will be constant. Your equation will give a velocity of zero, which means that you are using the wrong mathematical model. – David White Jul 24 '18 at 15:40

This is not a very easy to problem solve. Consider Newton's second law

$$m\ddot y = F_B - mg - b\dot y,$$ where $F_B$ is the buoyant force on the balloon. You already have to decide whether or not drag $\propto \dot y^2$ or drag $\propto\dot y$ is more appropriate for this situation. For slower speeds, which might be a good assumption, the latter will be fine, so I'll choose it for now. But I don't know how fast your balloon is.

We then know that the buoyant force is equal to the weight of the air we displace, but that the volume of the balloon changes over time:

$$F_B = \rho_{\rm air}gV(t),$$ where $V$ is the volume. We can derive $V(t)$ using the ideal gas equation of state:

$$V(t) = \frac{NkT(t)}{P(t)}.$$

Temperature $T$ is a function of time because it contains atmospheric gradients (e.g., lower temperature at high altitudes). Pressure is constant by Charles' law. By differentiating, we can see how the volume changes over time:

$$\dot V = \frac{Nk}{P}\dot T.$$

The problem is that $T(t)$ is not useful to us because we don't know how it changes over time. We can transform it using the chain rule, a variant of the $v\frac{dv}{dx}$ rule of classical mechanics. We find that

$$\dot T = \frac{dT}{dt} = \frac{dT}{dy}\frac{dy}{dt} = \dot y \frac{dT}{dy}.$$

Combining this with the last equation, we have our volume evolution

$$\dot V = \dot y \frac{Nk}{P} \frac{dT}{dy}.$$

Both in the upper stratosphere and in the troposphere, $T(y)$ has the form $T(y) = \alpha + \beta y$ (but not in the lower stratosphere, see this NASA link.)

Consequently we can see in those regions that the volume change is directly proportional to the speed

$$\dot V = \frac{Nk\beta}{P} \dot y,$$

but it is zero in the lower stratosphere. We can solve this differential equation to have

$$V(y) = \frac{Nk\beta}{P}y + V_0.$$

Now we can go back to Newton's second law

$$m\ddot y + b\dot y - \rho_{\rm air}g\left(\frac{Nk\beta}{P}y + V_0 \right) = mg;$$

or, more nicely,

$$m\ddot y + b\dot y - \left(\rho_{\rm air}g\frac{Nk\beta}{P}\right)y = mg + \rho_{air}gV_0.$$

$$\rightarrow \ddot y + b\dot y + cy = d.$$

This should have an over-dampened solution as described here.