How much force will my rocket experience on landing?

[mandbn]

I am trying to figure out how much force my rocket will experience on landing, so I need to know what the contact duration will be roughly when my rocket hits the ground, in the large grassy field it is being launched in in Texas. If this helps the launch site is in Stonewall, Texas and my 3481 gram rocket hits the ground at 5.1773 m/s.

 

[mandbn]

I am using 1/8 inch LOC plywood fins and I am worried they could break on impact.

 

[Steve Shannon]

Force = mass x acceleration
Acceleration = (change in velocity)/ time
Distance = average velocity x time
Average velocity while decelerating = 1/2 initial velocity.
Estimate the distance your rocket travels while decelerating (depth of grass?)
Solve for Force in terms of distance and you can calculate Force. Only simple algebra is needed.

 

[Bat-mite]

F = ma

 

[Bat-mite]

F = ma

Man, Steve got in a hair before me with a better post.

 

[Funkworks]

I think the range of possible or estimated impact durations is too wide to be useful. The range of calculated forces will be very wide as well.

Someone should fit pressure sensors on their fins and measure forces. Once we know a force, impact duration can be calculated with the posted equations (and/or mv = Ft, but even that is an approximation because force isn't constant over impact duration, and a true calculation requries calculus).

But: 5 m/s sounds rather slow, and 1/8" fins are common, and a broken fin can be repaired, and it would also depend on fin size, shape and count, so ... many other factors here.

 

[dr wogz]

The shape & span of the fins is also a concern..

Long swept back = possible damage: hits the ground first
trapezoidal / triangular above the end of the tube = much less likely to hit the ground first. (But beef up the aft end of the body tube)

 

[PhilC]

Funny, but I've been looking at the physics of impacts a lot recently. F=ma is a good start point, but its a bit more subtle than that. An equivalent form defines force as F=m(dv/dt) because acceleration is the rate of change of velocity.

The problem with calculating the force from impact is that we have no idea how long the impact lasts. We know the change in velocity dv because the rocket comes to rest from its descent speed. It's important to know how long that deceleration takes. As dt becomes shorter, the term dv/dt increases so the force increases. In the extreme case of the rocket coming instantly to rest, dt becomes zero and the force becomes infinite, which is clearly nonsense!

The only real way to measure the force on a rocket is to mount accelerators onboard and measure the deceleration over the time of the impact, a bit like crash test dummies. This would give a profile of how long the deceleration lasts and the profile of the acceleration, from which the average dv/dt can be estimated.

 

[Steve Shannon]

Funny, but I've been looking at the physics of impacts a lot recently. F=ma is a good start point, but its a bit more subtle than that. An equivalent form defines force as F=m(dv/dt) because acceleration is the rate of change of velocity.

The problem with calculating the force from impact is that we have no idea how long the impact lasts. We know the change in velocity dv because the rocket comes to rest from its descent speed. It's important to know how long that deceleration takes. As dt becomes shorter, the term dv/dt increases so the force increases. In the extreme case of the rocket coming instantly to rest, dt becomes zero and the force becomes infinite, which is clearly nonsense!

The only real way to measure the force on a rocket is to mount accelerators onboard and measure the deceleration over the time of the impact, a bit like crash test dummies. This would give a profile of how long the deceleration lasts and the profile of the acceleration, from which the average dv/dt can be estimated.

Rather than using time, put it into terms of distance and make a rough estimate of how far the rocket might penetrate or deform the surface it strikes. Or measure the distance afterwards if you want empirical data. No instrumentation necessary (although it would be nice ).

 

[PhilC]

Rather than using time, put it into terms of distance and make a rough estimate of how far the rocket might penetrate or deform the surface it strikes. Or measure the distance afterwards if you want empirical data. No instrumentation necessary (although it would be nice ).

That would give a good approximation in soft ground.

 

[AllDigital]

An 8 lb rocket landing at 17 fps should be just fine. I don't think more parachute will lower your risk much.

 

[boatgeek]

The big variable, as others have mentioned is your stopping distance. When I've landed on a wet grass field, my rockets with pointy fins often go 1-2" in. On concrete, they won't go in at all and all of the force goes into spinning the rocket around. That said, you're probably OK with reasonably strong fins at those speeds/weights. I would argue with your number of significant figures, though.

 

[JoePfeiffer]

Several other people have described the likeliest approach (start by estimating distance it will travel while stopping), and how hard that will be. I'll take another approach: why do you need to know this? What are you planning to use the information for, or what are you trying to do? Maybe there's a way to meet your real goals without the force estimate.

 

[BDB]

How much force? A lot...especially if you are landing on ice and your Chute Release freezes.

 

[Anthony Claiborne]

An accelerometer will provide you with the force of impact on the airframe. WHEN THE CHUTES HAVE FULLY OPENED, I have never seen anything more than about 2Gs of force on the airframe (when the chute has not opened, 10G is not especially unusual). Note though, that how that force is applied to the wing is determined by the attitude of the wing when it contacts the ground, and that may be more predictive of damage than the overall force of impact.

 

[BABAR]

I am trying to figure out how much force my rocket will experience on landing, so I need to know what the contact duration will be roughly when my rocket hits the ground, in the large grassy field it is being launched in in Texas. If this helps the launch site is in Stonewall, Texas and my 3481 gram rocket hits the ground at 5.1773 m/s.

As you can see, for your answer “it depends” on whether you land on a nice soft clump of grass or on the tennis court in the park (had that happen once, wasn’t pretty)

If your design allows it, having a forward sweep on the tail end of your fins or moving the fins forward a bit so the rocket lands on the casing rather than the fins can help a lot. The cost is a slight tax on performance, the less “far back” your fin surface is, the more fin surface you need to obtain the same level of stability.

for my scratch parachute and streamer rockets, I always use a forward sweep on the trail edge, so fins are protected. But I have a small field and am not an altitude seeker.

Then again, I am L0 and you or your sponsor must be high power certified (looks like this is a school project) given the mass of your rocket. The more experienced high power guys and gals here may be able to say whether it is better for a high power rocket to land on a fin or a motor casing, I don’t know how easy the casings are to break. Mine are low power cardboard or plastic single use motors, so I have no issues dinging the cases at all!

hope you get a great flight and a soft landing!

 

[JohnCoker]

It also really depends on the hardness of the surface. As a desert flier, I tend to bring 'em down slow, but I had one rocket break a fin a Snow Ranch because it came down with that fin hitting a rock.

 

[wonderboy]

Like they say: "It's not the fall that will kill you, it's the sudden stop at the end!"

 

[Zeus-cat]

I agree with AllDigital; you should be fine. However, you could hit a rock, land on really dry ground, land on a road or have a less than optimal chute deployment. Under most conditions landing at less than 20 feet per second is good.