The discussion centers on calculating the G-forces experienced by a model rocket upon impact with the ground at a terminal velocity of 31 mph (13.86 m/s). The rocket, falling from an apogee of approximately 1000 ft, is estimated to have embedded 3 inches (0.0762 m) into the ground, resulting in calculated accelerations ranging from 514 Gs to 128.4 Gs, depending on the method used. The calculations utilize the formula a = v² / 2s, where 'v' is the impact velocity and 's' is the penetration depth. The impact force is significant enough to raise concerns about the structural integrity of the rocket, especially if constructed from materials like cardboard and balsa wood.
PREREQUISITESPhysics students, hobbyist rocket builders, and anyone interested in the dynamics of impact forces and material resilience in model rocketry.
My calculations were for a solid (a steel ball, a rock or something similar). Cardboard would deform and probably be destroyed, but thereby creating a longer retardation distance for objects inside, thereby sort of protecting them.Zarich12 said:Thanks! One more question. If the that same rocket were made of cardboard and balsa wood fins, would it be worth getting back or would it be totally destroyed.
Try ##a=\frac{v^2}{2s}## instead.Svein said:You must tell us how deeply into the ground it ended up - then use the standard formulas (s = v0t and s = 0.5 at2) gives a = 2v02/s).
That figure of 0.005498 results from dividing 0.0762 meters by 13.86 meters/second. ##t=\frac{s}{v}##. But that assumes that the average velocity during the deceleration is equal to the initial impact velocity. That assumption is obviously wrong. A more reasonable assumption is constant deceleration, in which case the relevant velocity is the average of the initial impact velocity and the final velocity (rest). That is 6.93 meters per second (half of 13.86 m/s).Barbarian_Geek said:I got half the Gs of Svein using 0.005498 seconds to decelerate from 13.86 meters/second to zero in 0.0762 meters.