Impact force of model rocket with parachute

In summary: Materials fail from stress, (force per unit area) not necessarily force.Materials fail from stress, (force per unit area) not necessarily force.
  • #1
LT72884
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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
 
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  • #2
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.
 
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  • #3
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.
 
  • #4
LT72884 said:
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.
 
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  • #5
erobz said:
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
 
  • #6
LT72884 said:
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.
 
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  • #7
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
 
  • #8
LT72884 said:
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.
 
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  • #9
erobz said:
Yes, convert all figures into standard units.
ok, i get 156.2. Im guessing thats newtons?
 
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  • #10
erobz said:
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
 
  • #11
LT72884 said:
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.
 
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  • #12
erobz said:
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
 
  • #13
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.
 
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  • #14
LT72884 said:
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.
 
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  • #15
erobz said:
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
 

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  • #16
LT72884 said:
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.
 
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  • #17
LT72884 said:
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 ?

##\ ##
 
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  • #18
erobz said:
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.
 
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1. What is the impact force of a model rocket with a parachute?

The impact force of a model rocket with a parachute can vary depending on several factors such as the weight and speed of the rocket, the design and material of the parachute, and the surface it lands on. However, on average, the impact force can range from 5-20 pounds.

2. How is the impact force of a model rocket with a parachute calculated?

The impact force of a model rocket with a parachute can be calculated using the formula F=ma, where F is the impact force, m is the mass of the rocket, and a is the acceleration due to gravity. This formula can give an estimate of the impact force, but it may not be accurate due to other variables.

3. Can the impact force of a model rocket with a parachute be reduced?

Yes, the impact force of a model rocket with a parachute can be reduced by using a larger and more efficient parachute, decreasing the weight of the rocket, and launching it at a lower speed. Additionally, landing on a soft surface such as grass or sand can also help reduce the impact force.

4. Is the impact force of a model rocket with a parachute dangerous?

The impact force of a model rocket with a parachute is relatively low and is not considered dangerous. However, it is always important to follow safety precautions and launch the rocket in a safe and open area away from people and buildings.

5. How can the impact force of a model rocket with a parachute be measured?

The impact force of a model rocket with a parachute can be measured using a force sensor or a smartphone app that can measure acceleration. Another way is to use a high-speed camera and analyze the footage to calculate the impact force. However, these methods may not be accurate and should be used as estimates.

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