At what point does model rocketry have to morph into HPR?

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Steve Shannon said:
You’re correct that the kinetic energy is proportional to the square of the velocity, but the equation for kinetic energy at any of the velocities we’ll ever achieve is e = 1/2 m * v^2.
Conservation of energy and conservation of momentum must both be considered when analyzing the potential for damage.
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Steve, you beat me to this.
 
Rocketry is dangerous. Any other statement is silly. Dangerous doesn’t mean you shouldn’t do it. Just that it should be done with care. Follow the rules, and build proper recovery systems.

One of the reasons I dislike large, highly populated launches is that you’re covering more spots on the roulette tables, and you’re firing fistfuls of marbles at the wheel as fast as you can.
 
All else being equal, if my truck's getting hit I'd rather it be a couple of grams than a couple of pounds.
 
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lakeroadster said:
Velocity Squared... and yes... Really..
Click to expand...

Can't tell if you are messing with the folks on the forum or if you really haven't noticed that the "c" in that expression is the vacuum speed of light? E = mc2 is the expression for energy of a particle moving at relativistic speeds. c is a constant. It is not the speed of moving body. That speed is comprehended in the relativistic mass m. Rendered as E = moc2 it is understood to be the rest energy of a body with rest mass mo.

https://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

In any event, seriously and for real, the speed of the rocket falling through air will depend upon the mass. The damage done by the impacting rocket penetrating the thing with which it collides will depend upon the force per unit area during the collision. The force of the impact is the rate at which the momentum of the rocket changes during the collision -- momentum is the product of mass and velocity (speed, for collision constrained to one dimension).

Look here https://hyperphysics.phy-astr.gsu.edu/hbase/colcon.html#c1

If you scroll up to the post where I did the math, you'll see that the terminal speed of the rocket goes as the square-root of the mass. This means that the momentum of the rocket at terminal speed goes as the mass to the 3/2 power.

Which means, holding all other parameters equal (rockets of identical dimensions, each falling from high enough up to reach terminal speed) doubling the mass of the rocket will increase the impulse delivered at impact by about 2.8X. Assuming that the duration of the impact is the same, and that that the area of contact is the same, the 2X heavier rocket will exert a compressive stress on your skull 2.8 times greater than that exerted by lower mass rocket.

You can figure it in terms of energy, too. The work done by non-conservative forces is what brings the rocket to rest in your skull. Those non-conservative forces are the rigid forces holding your skull and brains together. So if the rocket has greater kinetic energy, there will be more energy available to scatter skull fragments and liquify brains.

The KE of the falling rocket, as Steve Shannon and others have pointed out, goes as the product of mass of the rocket and the square of the speed (which is denoted by "v", not "c"). There is also a factor of 1/2

https://hyperphysics.phy-astr.gsu.edu/hbase/ke.html#ke

Since terminal speed goes as the square root of mass, the KE of the falling rocket will go as the mass of the rocket to the 5/2 power. Which means, again for rockets of identical dimensions fallig at terminal speed, doubling the mass of the rocket will increase the kinetic energy delivered at impact by a factor of 5.7. that is to say, if you double the wight of the rocket -- holding all other dimensions the same -- you will increase the kinetic energy delivered by an impact at terminal speed by almost 6X/

Edit: Okay, that was embarrassing. Forgot how to square the square root. Kind of knocks the bottom out of all my highfalutin physics arguments.

After a moment's consideration, I have decided that the square of the square root of m, multiplied by m, is probably m2 -- not m2.5. Accordingly, I revise my previous: Which means, again for rockets of identical dimensions falling at terminal speed, doubling the mass of the rocket will increase the kinetic energy delivered at impact by a factor of 4. That is to say, if you double the weight of the rocket -- holding all other dimensions the same -- you will increase the kinetic energy delivered by an impact at terminal speed by 4X

But really, it makes more sense -- in terms of the OP's concern -- to think about in terms of the impulse (change in momentum) delivered by the impactor.

Now, go ahead, tell met that is not the volts but the amps that make electricity dangerous. I've got a whole ranty bit I can unload on you if you want to try that one on.
 
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I'd say the Mach number for a minimum diameter build especially a HPR build makes it more dangerous inherently by kinetic energy. You'll start having more energy on paper than a 20mm-37mm AAA cannon from WW2 on paper. There's a good reason why we all do the sky is clear and it flies at sanctioned waiver event into clear sky. You can snicker and snot all you want but I've had L-1 multistages CF scratch MD with 137,000 Joules of kinetic energy on paper on an I to H motor. I called a glue company Aeropoxy told them lives were on the line; got rejected by a engineer over thermal concerns then bought cotronics4700 epoxy. The total temperature calc by advanced fluids confirmed the space epoxy was slightly overkill but money well spent for M1.7. Granted that was uni launch. Man I got trolled here hard. It was only pulling 168 G. We would've melted other low temp epoxies and it wasn't a kit you could buy we developed it from scratch. Then NAR stuck our tiny itty bitty yardstick rocket way out away from people on a salt flat for a student launch. But HPR starts where the safety codes and specific impulse of the H motor starts that qualify for cert. Being new and unskilled makes it even more risker. There's even MPR G motors that are L-1 required to buy those are no joke in a 24mm MD and I doubt the cardboard would work. Some of those are going to go Mach 1.37 and 8000 ft. Yeah I understand a big diameter draggy subsonic rocket works on plywood and cardboard up to M it has more drag force by larger area and Cd. But that same situation in a MD with less area and Cd shreds.

I'm guessing once you get into serious K or L motors in MD the thermal requirements need those aluminum tips as heat sinks by Mach and temperature when you really get over Mach 2.5-3. You can always pick an epoxy stronger than you need it, I won't laugh. Many here will. It'll give you extra margin of error on angle of attack in fin sim for bending stress and having a datasheet is nice in my opinion that a product meets known standards. You can peel fiberglass fins off some kits off at subsonic speeds when their span is too high for thickness and it flutters in FinSim. Everyone gets serious about the complex rockets. They care more to check over your work and they ask more questions. Just a noob rambling.
 
jlabrasca said:
Can't tell if you are messing with the folks on the forum or if you really haven't noticed that the "c" in that expression is the vacuum speed of light? E = mc2 is the expression for energy of a particle moving at relativistic speeds.
Click to expand...

Yes, I agree, wrong equation, Steve set me straight on that....

Steve Shannon said:
You’re correct that the kinetic energy is proportional to the square of the velocity, but the equation for kinetic energy at any of the velocities we’ll ever achieve is e = 1/2 m * v^2.
Conservation of energy and conservation of momentum must both be considered when analyzing the potential for damage.
Click to expand...

As I stated in my response to Steve...

lakeroadster said:
Thanks for that Steve. I guess the point I was making is since the velocity is squared, velocities impact on energy is much greater than mass.
Click to expand...

I merely made this post

lakeroadster said:
Velocity Squared... and yes... Really..
Click to expand...

To provide a link to an article that shows "Theory of Special Relativity" is indeed a "real" thing, in response to jimn post of:

jimn said:
Momentum equals mass times velocity. The special theory of relatively? Really?
Click to expand...

My head hurts.. I need a flow chart.. :facepalm:

I should know better than to weigh in on such matters... my apologies.
 
lakeroadster said:
My head hurts.. I need a flow chart.. :facepalm:

I should know better than to weigh in on such matters... my apologies.
Click to expand...
Physics. Think Kinetic energy, potential energy, and projectile motion. All that algebra crap you might have snoozed through during highschool. It seems complex until you break it down into pieces and understand the concepts. Dynamics comes later. The units may look similar but the concepts can be wildly different from other formulas which is the danger when you haven't practiced with or are unfamiliar with the ideas.
 
A bullet fired from a rifle has kinetic energy because it's a mass moving at a velocity, KE=0.5*m*v^2. The height of the barrel from the ground if the shooter simultaneously drops a lead bullet to ground has potential energy =m*g*h. The bullet represents a complex system of kinetic and potential energy further explained as mechanical energy a combination of PE and KE in a system. That's an analogy I use. If you want to see what bullet does to swing pendulum that involves dynamics course from engineering. Do not confuse with speed of light stuff energy or people will laugh.
 
During boost at burnout the rocket has maximum kinetic energy. During apogee at maximum height assuming the velocity is zero it will have maximum potential energy.
 
lakeroadster said:
I'm Your Huckleberry
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Sorry if I came off as mean. Some people have not had the opportunity to learn or maybe it's been decades and they just don't remember. Everyone has all sorts of talents in life.
 
Andrew_ASC said:
Sorry if I came off as mean. Some people have not had the opportunity to learn or maybe it's been decades and they just don't remember. Everyone has all sorts of talents in life.
Click to expand...
And some people need to take their meds regularly, and learn to write sparingly,and on-topic

Please go and take care of the first item ASAP !

*sigh*
 
lakeroadster said:
... provide a link to an article that shows "Theory of Special Relativity" is indeed a "real" thing, in response to jimn
Click to expand...

Yeah, I think you misread that "really". Pretty sure it was meant as an editorial comment, rather than an expression of doubt regarding the validity of special relativity.

lakeroadster said:
My head hurts.. I need a flow chart.. :facepalm:

I should know better than to weigh in on such matters... my apologies.
Click to expand...

No apologies necessary -- except maybe from me for being a bit of an old FUSSPOT. Unless you plan to certify L2, you really don't need to bother yourself with differential equations.
 
I didn't got through all the posts, but it seems some think that mass/weight affects terminal velocity.

I'm not a scientist or mathematician, but NASA has lots of both. If you go to the NASA site, you can use their tool to calculate the terminal velocity of any object on any planet in the solar system. You only need to enter three things;
  • The planet you're on
  • The altitude you want
  • The Coefficient of Drag for your object
Nowhere does it ask for mass or weight.

That tells me that rhildinger is correct that two objects with the same Cd will reach the same terminal velocity irregardless of the mass.

Now the amount of energy each has is directly related to the mass.
 
You mean if there's very little mass it doesn't matter how fast it's going...

* gets peppered with photons *

... and if there's enough mass it doesn't matter how slow it's going?

* gets smeared by an iceberg *
 
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Handeman said:
I didn't got through all the posts, but it seems some think that mass/weight affects terminal velocity.

I'm not a scientist or mathematician, but NASA has lots of both. If you go to the NASA site, you can use their tool to calculate the terminal velocity of any object on any planet in the solar system. You only need to enter three things;
  • The planet you're on
  • The altitude you want
  • The Coefficient of Drag for your object
Nowhere does it ask for mass or weight.

That tells me that rhildinger is correct that two objects with the same Cd will reach the same terminal velocity irregardless of the mass.

Now the amount of energy each has is directly related to the mass.
Click to expand...

link please

like this one

https://www.grc.nasa.gov/www/k-12/airplane/termv.html

termv.gif
 
dhbarr said:
You mean if there's very little mass it doesn't matter how face it's going...

* gets peppered with photons *

... and if there's enough mass it doesn't matter how slow it's going?

* gets smeared by an iceberg *
Click to expand...

No, I'm not talking about near light speed objects in space or rogue planets colliding. I'm talking 0 to about 300 lb object falling through earth's atmosphere between about 0 and 5,000 ft above sea level.

Not that the others aren't interesting.....
 
Handeman said:
No, I'm not talking about near light speed objects in space or rogue planets colliding. I'm talking 0 to about 300 lb object falling through earth's atmosphere between about 0 and 5,000 ft above sea level.

Not that the others aren't interesting.....
Click to expand...

I get that.

I was absurdly exaggerating for effect. This is where you take a concept to one or more logical extremes, such as a scale ranging from only-relative-mass to mountain-floating-by.

Expressed differently: for comedy purposes, but also to help illustrate the concept, I pushed the mass-vs-velocity sliders to their limits.

Then I murdered the already-thin joke twice, and now here we are.
 
There is nothing wrong with the discussion IMHO. It just appears that the majority had less of an understanding of terminal velocity than I at least expected. Those working with rockets should understand at least the basic concepts even if they don't know the math off the top of their heads. At least, IMHO.

To make matters more interesting, Cd is definitely not a constant. I do enough work in low Reynolds numbers aerodynamics where practically nothing is a constant. At the low end (slow x small) drag coefficient changes fast as the flow characteristics change. Approaching Mach and going faster it also changes fast. Essentially if the flow characteristics stay close to constant the drag coefficient will say close to constant. But the turbulent transition point is going to move based on Reynolds number (size and velocity related) so it is speed dependent. Be high enough and aero enough and terminal velocity at that altitude might be supersonic (to add to the fun, whether or not terminal velocity is supersonic can depend on whether the starting velocity is sufficiently supersonic). The drag coefficient will be different than it would be at, say, Mach 0.5. If it were not different, then breaking Mach for early jets would have been no big deal. The flow changes fast around Mach, and therefore so does the drag coefficient.

https://sbainvent.com/fluid-mechanics/pictures/mach-supersonic-drag.jpg

In Physics and Engineering, there are rules of thumb, standard theories and methods which work most of the time, and advanced theories and methods to handle the edge cases you don't run into all the time. It is important to know which to apply to a problem!

Gerald
 
This thread has a good working base and several knowledgeable folks with ready access to textbooks: think we could come up with a quick three-paragraph blurb about when rockets fail at falling for publication in one of The Usual Places?
 
Any resource that can make the physics of the hobby more accessible to the general population is a good thing in my book.
 
Buckeye said:
Huh? Did the L2 exam suddenly get meatier while I wasn't looking?
Click to expand...

Lol, of course not. I don't think DEq, are ever "needed" at the basic sport level, but they certainly help if you get into interesting control systems or want to try and develop simulations.

That being said, there are a ton of mechanics and basic physics that could be beneficial to the average flyer. Maybe have less "I do this because he said it worked for him" moments

Unless you were joking, in which case, yogot me
 
Nytrunner said:
Lol, of course not. I don't think DEq, are ever "needed" at the basic sport level, but they certainly help if you get into interesting control systems or want to try and develop simulations.

That being said, there are a ton of mechanics and basic physics that could be beneficial to the average flyer. Maybe have less "I do this because he said it worked for him" moments

Unless you were joking, in which case, yogot me
Click to expand...

Not joking. I was asking the poster why he thinks DEq are important for L2 certification. 99% of L2 (and L3 ) flyers could not identify, apply, or let alone solve, a differential equation. Sure, I used to write simulation codes, but now the popular software do all the setup and solving for me. And why L2? The same DEq's apply to all aspects of rocketry.

That said, I think the L2 exam could use some beefing up to include questions about simulations and interpreting flight data. In its current form, the NAR exam is mostly a dated exercise in useless memorization of distance tables and impulse ranges. "Titanium sponge?" Who even uses that term?

With all the newbies racing through the certifications, are the required tests of skill doing the job? Eventually, some ninny with no clue of flight trajectory or the deployment logic of his altimeter is going to do some serious damage.
 
Buckeye said:
... In its current form, the NAR exam is mostly a dated exercise in useless memorization of distance tables and impulse ranges...
Click to expand...

Geez. Why don't you go ahead and post the questions on the internet while you're at it?

It is worrisome that there are so many fliers who don't have a strong(er) background in physics. You should need to solve some differentials equations for the L2 cert.

For that matter, you should probably be required to work a few variational calc. problems to get a driver's license or a voter registration card (no, that's elitist -- suffrage is, and ought to be, universal. Which, in turn, is why an education in mathematics up to differential geometry and abstract algebra ought to be universal and compulsory)

How will you teach physics to folks who just want to fly big rockets? How much physics do you need?

Here's a math problem: What is the carrying capacity for the innumerate in any population?
 
jlabrasca said:
Geez. Why don't you go ahead and post the questions on the internet while you're at it?

It is worrisome that there are so many fliers who don't have a strong(er) background in physics. You should need to solve some differentials equations for the L2 cert.

For that matter, you should probably be required to work a few variational calc. problems to get a driver's license or a voter registration card (no, that's elitist -- suffrage is, and ought to be, universal. Which, in turn, is why an education in mathematics up to differential geometry and abstract algebra ought to be universal and compulsory)

How will you teach physics to folks who just want to fly big rockets? How much physics do you need?

Here's a math problem: What is the carrying capacity for the innumerate in any population?
Click to expand...
I know your statement is ment as sarcasm, but some may miss that fact and get bent out of shape....
 

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