Impact force of model rocket with parachute

[LT72884]

Homework Statement

NONE

Relevant Equations

NONE

I just launched a 2.2kg model rocket that stands about 5 feet tall and has a airframe that is 4 inches in diameter. I am trying to find how much the impact force is when the rocket hits the ground at 16MPH.

Whats the best approach? Momentum? KE? or something else:)

thanks

 

[kuruman]

The best approach IMO is to use the equation $F=\frac{mv}{\Delta t}$ which gives you the average force. Of course you have to know (or estimate) the time $\Delta t$ needed for the rocket to come to rest from 16 mph. That time is longer for ground that "gives", e.g. mud, than ground that does not, e.g. concrete.

 

[LT72884]

not sure i follow on the time piece? once it hits the hard ground, it is stopped immediately? Or do you mean how long was it falling at 16MPH? if so, then that was 180 seconds from the time the chute deploys to the time it hits the ground.

 

[erobz]

not sure i follow on the time piece? once it hits the hard ground, it is stopped immediately? Or do you mean how long was it falling at 16MPH? if so, then that was 180 seconds from the time the chute deploys to the time it hits the ground.

They mean the duration of the impact. The total time it takes for the rocket to go from 16 mph to 0 mph. If it is "instant", then your rocket explodes on impact. If the ground doesn't deform, then your rocket does...which is what you don't want.

 

[LT72884]

They mean the duration of the impact. The total time it takes for the rocket to go from 16 mph to 0 mph. If it is "instant", then your rocket explodes on impact. If the ground doesn't deform, then your rocket does...which is what you don't want.

ok, so what do i put for the number then because the time will be super super small

 

[erobz]

ok, so what do i put for the number then because the time will be super super small

That is not an easy question to answer. You might try to model the ground and/or rocket as colliding damped springs.

 

[LT72884]

well, i will use 0.1 seconds.
force = (2.2kg)(25.75KmPH)/0.1

should i convert the Km per hour to meters per second?

thanks

 

[erobz]

well, i will use 0.1 seconds.
force = (2.2kg)(25.75KmPH)/0.1

should i convert the Km per hour to meters per second?

thanks

Yes, convert all figures into standard units.

 

[LT72884]

Yes, convert all figures into standard units.

ok, i get 156.2. Im guessing thats newtons?

 

[LT72884]

Yes, convert all figures into standard units.

hmm, that doesnt seem right. 156 newtons is about 35 lbs force acting on the fins when it hits the ground. That seems way to high. That would snap them. If i make the time faster, the newtons increases. so something is not correct. or maybe it really is 35Lbs force of the area of all 3 fins....

thanks

 

[erobz]

hmm, that doesnt seem right. 156 newtons is about 35 lbs force acting on the fins when it hits the ground. That seems way to high. That would snap them. If i make the time faster, the newtons increases. so something is not correct. or maybe it really is 35Lbs force of the area of all 3 fins....

thanks

Materials fail from stress, (force per unit area) not necessarily force.

 

[LT72884]

Materials fail from stress, (force per unit area) not necessarily force.

true, just my brain has a hard time seeing how a 35LB weight falling on my 3 fins wouldnt damage them.. But it would all be in compression anyway. I think it would hurt to have a 35 lb weight fall on me hahahaha

 

[erobz]

They very well might shear off of the hull. In reality they don't land level, and it could potentially be much worse. Then again, pulling 0.1 s out of thin air might not be accurately capturing the force.

 

[kuruman]

ok, so what do i put for the number then because the time will be super super small

It's a number that it is best to measure to eliminate guessing. As I already mentioned, the stopping time depends on where the rocket lands. Smartphones have accelerometers and apps to access them. If you get a recording of the acceleration as a function of time, you have the force as a function of time.

 

[LT72884]

They very well might shear off of the hull. In reality they don't land level, and it could potentially be much worse. Then again, pulling 0.1 s out of thin air might not be accurately capturing the force.

yeah, i have my fins mounted through the hull and on to the motor mount tube. The MMT is then epoxied into place using 3 rings. Each ring is 1/4 inch thick and 4 inch in diameter.
The fins mount in between the wooden rings. if that makes sense

 

[erobz]

yeah, i have my fins mounted through the hull and on to the motor mount tube. The MMT is then epoxied into place using 3 rings. Each ring is 1/4 inch thick and 4 inch in diameter.
The fins mount in between the wooden rings. if that makes sense

So, why not put a larger parachute to slow it down further?

I imagine if anything yields in that setup, its going to be the cardboard body.

 

[BvU]

true, just my brain has a hard time seeing how a 35LB weight falling on my 3 fins wouldnt damage them.. But it would all be in compression anyway. I think it would hurt to have a 35 lb weight fall on me hahahaha

That's because you imagine it falling from some substantial height ?

What you calculated is the force that would be exerted on the ground if the 35 lb would rest there

Shock dampers ?

$\$

 

[LT72884]

So, why not put a larger parachute to slow it down further?

I imagine if anything yields in that setup, its going to be the cardboard body.

it has a 4 foot parachute. Thats the biggest i can use or it will not fit in the payload section of the rocket. I hit 5300 feet though so i am happy with that haha. I am now certified to launch high powered rockets.