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  • With a light weight of only 60g, achieve speeds of up to 312 mm/s!
  • Use its throttle valve to control the amount of air released and the speed/duration of a run. Use the rudders to control direction.
  • Adjustable platform for an optimal balance in the water and the minimization of drag area on the floaters.
  • No batteries needed, only balloons and the power of your lungs.


In a similar way that air is used to push the strider through water, airplanes get propelled through air. This type of propulsive force is known as THRUST. Its computation is very important in the aeronautics world and we think you would like to know how it is achieved.

Let’s take a look first at what is happening. The balloons apply a force to the air in the form of pressure and force it to exit. Even though the air does exit, it resists movement and applies a force back to the balloons equal in magnitude but in the opposite direction. An action is followed by a reaction. This is Newton’s third law! Since the balloons are attached to the strider, the air’s reaction is effectively transferred on to it and this is what causes it to move in the opposite direction. Action and reaction are equal in magnitude so computing the force exerted on the air (action) will tell us how much thrust is applied to the strider (reaction).

To quantify the force exerted on the air you need to understand Newton’s second law and, for this you need to first be introduced to the concept of “momentum”. A body of mass (m) moving at a certain velocity (v) has a momentum equal to the product of these two (mv). So we can increase the momentum of an object by either increasing its speed or increasing its mass. Think of a snowball rolling down the mountain. As it rolls it grows larger in mass and so its momentum increases (it is also speeding up…).

In its purest form, Newton’s second law states that a force is needed to make an object change its momentum over time. The value of such force is equal to the change of momentum desired over the time desired. So if we want a resting tennis ball to change its velocity to 5 m/s and achieve this in a process lasting 1 second, we will have to exert 10 times more force than if we wanted it to acquire the same velocity after 10 seconds.

F = \dfrac{d(mv)}{dt}\approx \dfrac{\Delta(mv)}{\Delta t}
F = force [N or kgf], m = mass [kg] , v = velocity [m/s], t = time [s]

When the object moving is a solid, it is very easy to think of the change in momentum as the product of the mass and the acceleration (hence the F=ma). But when it’s a fluid like air, we take the “snowball” approach. Sitting outside the strider, we see air mass exiting the strider at a certain velocity being increasingly delivered to the atmosphere. So the change in momentum of the air can be related to the change in mass moving at a given velocity. This change of mass over a given period of time is what is known as mass flow. There is a flow of mass of air exiting the balloons of the strider, there is a flow of mass of air reincorporating to the atmosphere… If we think that the velocity of the air exiting the strider does not change with time during deflation (not an accurate approximation for our balloons but certainly for an aircraft engine!) we can re-write Newton’s second law for thrust computation:

 T = \dfrac{\Delta(mV)}{\Delta t} \approx V \dfrac{\Delta m}{\Delta t}\approx V\dot{m}
T = Thrust [N or kgf], ṁ = mass flow [kg/s]

All we need now is to know the exit velocity of the air and compute the flow of air mass. We can know how much air volume flows out of the strider if we know the area of the exhaust port and the air velocity (See the figure) . The product of the volume flow and the density equals the mass flow.

 \dot{m} =\rho\dot{V_{ol}}=\rho A V
ρ = Density of air [kg/m3], A = cross section area [m2]

So given that we know rho = 1.275 kg/m3 and the exit port of the stride is 32mm2, really the key thing to know is the exit velocity of the air. Using two balloons of 200mm of diameter we have measured the exit velocity through the valve exhaust port fully opened. It takes about 10s to deflate. See the graph we have obtained

In rough terms we can then estimate that the force on the strider using two balloons is 0.002N or 19.2 gf (grams-force, nearly the same as the weight of AA battery). Perhaps not a lot, but definitely enough to move the strider. If no other forces would be acting on the strider, by the end of the 10 s period that it takes for the balloons to deflate, the velocity of the strider would be 312 mm/s:

 T = m_{strider}\dfrac{\Delta V}{\Delta t} \longrightarrow V_f = \dfrac{T \Delta t}{m_{strider}} + V_o
Vf = final velocity [m/s], Vo = initial velocy = 0

Unfortunately, Thrust is not the only force acting on the strider movement. There is Gravity, Buoyancy and Drag. Drag is a force due to the friction between the strider body and the water. It depends on the area of the cross section of the floaters that is immersed in water. Although we won’t try to compute Drag here, we can say that your strider will travel faster if less cross sectional area is exposed to the water. More reason then, to try to balance it properly . What else could be subject to drag? Are the balloons offering a large cross section area against wind? How would air drag compare to water drag on the body?

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