Wild Rocket: Newsletter Challenge (01/05/10)

Vertical = 2.5 x 9.81 x 8² = 1569.6m

Slowdown = 9.81 x s² = 1569.6 m/s

s² = √( 1569.6/9.81) = 160.

s = 12.649

Total Distance = 1569.6 + 9.81 x 12.649² = 1569.6 x 2 = 3139.2 m

Max height reached = 3139.2 m

Regards JD

Did you do a sanity check on your answer? 3 km altitude for a model rocket? You could be shooting down small planes!

With force specified, not acceleration, doesn't the acceleration (over time) depend on the mass of the propellant vs. the empty rocket mass? With a constant force, the acceleration should increase over time, at least if atmospheric drag is ignored. In such a case, none of the questions can be answered without making some additional assumptions.

-J

That's true, but without a fuel mass and burn rate the rocket equations can't be used. A typical 20 N-s (2.5N x 8s) model rocket motor has about 25 g of fuel, so this could be used to determine an average burn rate of 0.003125 kg/s and then plugged into the rocket equations. Using the same methodology as the rocket challenge from a few months ago, the equations are:

v(t) = -800 ln(1.0 - 0.003125t) + C

s(t) = -256000 (1.0 - 0.003125t) (ln(1.0-0.003125t) - 1) + C

which after 8 s are:

v(8) - v(0) = 20.25 m/s

s(8) - s(0) = 80.68 m

These are about 1% higher than my solution below using straight dynamics, so the effects of the changing mass are fairly negligle in this case. However, as the burned fuel mass becomes a larger portion of the rocket's initial mass, the error will become much larger and the rocket equations must be used.

A typical 20 N-s (2.5N x 8s) model rocket motor has about 25 g of fuel, so this could be used to determine an average burn rate of 0.003125 kg/s and then plugged into the rocket equations.

However, 2.5N gross thrust is not enough to lift the rocket.

One small change. The acceleration is still 2.5m/s^2 the instant the fuel runs out so the the deceleration should be -7.3m/s^2 (-9.8+2.5) which gives 27.4 meters extra or a total of 107.4 meters traveled.

At the instant immediately after the rocket flames out, the acceleration is goes to -9.81 m/s². At that instant, the rocket begins to decelerate as calculated by DAC (the 2.5N is history).

(This assumes that the original rocket thrust is 12.31N (for a net thrust of 2.5 N). Otherwise the rocket would not take off.)

You are correct! I was confusing velocity with acceleration. The acceleration goes to -9.81 at flame out.

Once the fuel has been spent, the upward acceleration of +2.5 m/s^2 is no longer realized and the only Force on the Rocket is that of Gravity. The original result is closer to correct.

As fuel burns, the mass of the rocket will decrease, and the rate of acceleration will increase. We need more information in order to calculate the rate of decrease in mass, such as either the initial mass of fuel or perhaps the exhaust velocity. The gross thrust of 2.5N + 9.8N is not enough info.

s= ut+1/2at²

= 20*2.04 + 1/2(-9.81) 2.04²

= 20.39

Answer is correct but the method is wrong

Following your unfair critique, I'm going to add to DAC's tally of GAs.
His method is merely a step more fundamental than yours - though I would agree that the changing directions of the coordinates are not stated.

Doesn't this:

20*2.04 + 1/2(-9.81) 2.04²

evaluate to:

-20.39?

I think I prefer DAC's method.

My model rocket spreadsheet also crashes (along with the rocket) when it tries to calculate the square root of numbers 0 or below, which appears to happen every time the weight of the rocket is to great for the impulse and thrust of the motor.

I am thinking we need to reread the challenge. The height of the rocket will not change, unless the motor "CATO's" and burns the rocket to the ground.

Maybe we could calculate how high the motor went without the rocket.

CATO example

dac1267's analysis is correct, assuming that 2.5N is the net force, meaning the engine really has 9.8N (to balance gravity force on 1.0 Kg) + 2.5N (accelerating force) = 12.3N of total force. The rocket will travel 80m during the burn, per dac's equation of motion for distance travelled during steady acceleration.

We have to assume air friction to be zero; otherwise this turns nasty, involving some higher-level math, since air friction varies with speed and shape of the object.

An alternate (leading to same result) analysis after firing ends is to say that, by conservation of energy:

Potential Energy(=mgh) gained by the top of the climb is equal to Kinetic Energy(=1/2 mv²), achieved during firing, leading to h = v²/2g = (20m/sec)x(20m/sec)/2/9.8m/sec² = 20.41m

Adding the two distances gives the 100.41m total height.

This is a Grade 11 physics question type.

I do not believe that it can be assumed that any object "carrying" it's own fuel accelerates at a constant rate. It is the momentum of the exhaust gases that power the rocket flight. See http://ocw.mit.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/77BBA399-5FAB-4A16-AA53-29A30382EEB2/0/supplement8.pdf

This is a Grade 11 physics question type.

Or grade 9 in the developed countries.

Regards JD.

Yikes! Guest is right - the answers are all zero!

if this is a typical rocket with 10% payload, it starts at 1 KG and ends up in 8 seconds at end of burn at 100 grams.

It starts at 2.5G and at the end accelerates at 25G

so if we average it out to 13.5 G and 550 grams for 8 seconds.

This little whizbang will be going pretty quick?

But the net force is negative: 9.81 newtons down, due to gravity, 2.5 newtons up due to thrust. Therefore, the rocket never moves, unless at some point late in the burn the mass of rocket and fuel is less than 254 grams. But that is unlikely in a model rocket -- their fuel is not such a high percentage of total mass.

I wonder if something went haywire in formulating the question.

The newton is the unit of force derived in the SI system; it is equal to the amount of force required to accelerate a mass of one kilogram at a rate of one metre per second per second. In dimensional analysis, F=ma, multiplying m (kg) by a (m/s²), the dimension for 1 newton unit is therefore:^[1]

[edit] Examples

• 1 N is the force of Earth's gravity on an object with a mass of about 102 g (¹⁄_9.8 kg) (such as a small apple).
• On Earth's surface, a mass of 1 kg exerts a force of approximately 9.80665 N [down] (or 1 kgf). The approximation of 1 kg corresponding to 10 N is sometimes used as a rule of thumb in everyday life and in engineering.
• The force of Earth's gravity on a human being with a mass of 70 kg is approximately 686 N.
• The dot product of force and distance is mechanical work. Thus, in SI units, a force of 1 N exerted over a distance of 1 m is 1 N·m of work. The Work-Energy Theorem states that the work done on a body is equal to the change in energy of the body. 1 N·m = 1 J (joule), the SI unit of energy.
• It is common to see forces expressed in kilonewtons or kN, where 1 kN = 1 000 N.

Correct.

Thus, a 1kg rocket exerts a force of 9.81 N against the launch pad. 2.5 N of upward force changes this to a net force of 7.31 down. In other words, the rocket would accelerate downward, if the launch pad were pulled out from under it. With one finger, you can exert a force of 2.5N, and doing so under a 1kg rocket would not lift it.

The part of your quote that applies:

On Earth's surface, a mass of 1 kg exerts a force of approximately 9.80665 N [down] (or 1 kgf). The approximation of 1 kg corresponding to 10 N is sometimes used as a rule of thumb in everyday life and in engineering.

Do to the phrasing of the question, I believe they intend that the rocket exerted 12.32 N of thrust against gravity's 9.82 N of attraction. Thus, "ASCENDS subjected to a force of 2.5 Newtons"

Yep. Initial velocity is zero. As propellant burns, the weight might decrease enough to lift off, but there is insufficient information to determine if that will happen and how much burn time would remain to calculate a non-trivial trajectory.

Rocketguy

GA guest, maybe it's a 0.1 kg rocket (100 grams)

This is classic example of conservation of momentum You need to know mass of the exhaust gas during the 8 sec engine burn. See Paul Tipler's Physics for scientists and engineers, Fourth Edition Volume 1 8-8 Rocket Propulsion The final equation for velocity at the end of the engine burn is v= Thurst*ln(mass after fuel used)/(initial mass) - g*burn time

First term should be the minus velocity of exhaust gas not thrust

I agree with guest in post #7

my spreedsheet from model rocket days (based on Barrowman's equations) indicate the rocket does not get off the pad

Even with making assumptions on rocket area, propellant mass,.......etc.

As long as the rocket weight exceeds 2.5N it will not lift off. If the fuel mass is greater than 3/4 kilogram then there comes a time during the burn that the acceleration force exceeds the remaining weight of the rocket. But without knowledge of the ratio of fuel mass to rocket mass it is not possible to determine the height of ascent.

Disregarding the opposing force due to air friction, the gravity force is G=m*g=1Kg*9.81m/sq.s=9.81N which is way bigger than the propulsion force of 2.5N. The rocket will not go anywhere. Not this time.

Not being an engineer and not possesing the education and math skills needed to answer the speed and distance portions of the question, I would like to comment on the discussion of the rocket leaving the pad. The question specificially states that the rocket ascends and that it travels. This would indicate to me that the 2.5 newtons is the net force resulting after the force of gravity has been overcome.

The question also states that the rocket continues upward until its velocity is zero. This also indicates that the rocket has left the pad.

I would have to agree with those calculating the effect of the 2.5 newtons as net after gravity due to the wording of the question.

In order to make the question more interesting let's just assume it left the pad as per the previous post.

M. S.

It's a dud! "Kaplunkit"

LOL

Since the problem is boring if it is only a 2.5 N force, let's assume it's a net 2.5 N upward force (12.3N propulsion - 9.8N gravity).

The rocket will accelerate at (2.5 N / 1 kg) * 9.8 m/s2 = 24.5 m/s2

The thrust will end at 0.5 * 24.5 * 8 * 8 = 784 m

It will be traveling at 24.5 m/s2 * 8 s = 196 m/s

It will continue for 196 / 9.8 = 20 more seconds

The additional distance travelled will be (196 * 20) + (0.5 * -9.8 * 20 * 20) = 1960m for a total height of 2744 m

Since the problem is boring if it is only a 2.5 N force...

Aren't the ones that fizzle more exciting?

The rocket will accelerate at (2.5 N / 1 kg) * 9.8 m/s2 = 24.5 m/s2

Hmmm... F=MA.

For a reality check, you would expect that the acceleration would be about 1/4 g, i.e around 2.5 m/s². Your figure of 24.5 m/s² is off by roughly an order of magnitude, because your formula suggests that F/M * A = A

I am assuming a 2.5N as the net force to propell the rocket upward.

Applying equations of motion

v=u+at

s=ut+1/2at^2

v^2=u^2+2as

Stage 1

u=0 m/s

t=t1=8 sec

As F=ma

2.5=1xa

a=2.5m/sec^2

applying v=u+at

that is v=0+2.5(8)=20m/sec this is the speed at 8 sec at fuel out condition

Applying v^2=u^2+2as

that is 20^2=0+2x2.5s, hence s=s1=80m this is the height the rocket reached before fuel out.

Stage 2

u=v1=20m/sec

v=v2=0m/sec

a=-g=-9.81m/sec^2

s=s2

Applying v^2=u^2+2as

which is 0=20^2+2(-9.81)s

hence s=s2=20.4m

hence total distance travelled = s1+s2= 80+20=100m.

QED. Cheers

This is very similar to the "Rocket Science" challenge from a few months ago, where the official answer was that the rocket never did get off the ground due to insufficient thrust - I can't image they would give us the same problem again so soon. This challenge states that the rocket ascends with a force of 2.5 N, so I take that to mean that the 2.5 N is the net force acting on the rocket. This means that the engine is actually providing 2.5 N + (1 kg)(9.8 m/s²) = 12.3 N of thrust.

This thrust requires more power than a 20 N-s engine can deliver (as I stated in post #6). In actuality the rocket needs (8 s)(12.3 N) = 98.4 N-s of impulse power to meet the challenge requirement. I couldn't find a 100 N-s model rocket engine, but this could possibly be accomplished with 5 stages of 20 N-s engines. These would fit the challenge too since each engine only burns for 1.6 sec (8 s total burn). However, this means the fuel load would be about 125 g (25 g/engine) so the rocket equations should be used to determine velocity & height. For this problem, the engine thrust is assumed to be constant, even though the burn profile for these engines is not constant (thrust is highest during first 0.5 s then falls off). However the calculated results should be pretty close.

Time dependent acceleration:

Average burn rate = 0.125 kg / 8 s = 0.0156 kg/s

a(t) = F_net/m(t) = 2.5 / (1 - 0.0156t)

Integrate acceleration to get velocity:

v(t) = -160 ln(1 - 0.0156t) + C

v(0) = 0 → C = 0

v(t) = -160 ln(1-0.0156t)

Integrate velocity to get position:

s(t) = 10240 (1-0.0156t)(ln(1-.0156t) - 1) + C

s(0) = 0 → C = 10240

s(t) = 10240 ((1-0.0156t)(ln(1-0.0156t)-1) + 1)

Solving for v & s at t=8 s (when rocket is out of fuel):

v(8) = 21.4 m/s

s(8) = 83.6 m

At this point, regular dynamics take over.

Time to reach zero velocity:

t = v / a = 21.4 / 9.81 = 2.2 s

Travel distance during coast:

s = ½ a t² = (0.5)(9.81)(2.2)² = 23.3 m

So assuming a straight vertical trajectory, the height when the rocket runs out of fuel is 83.6 m, the velocity at that point is 21.4 m/s and the maximum height achieved is 83.6 +23.3 = 106.9 m.

This is one of those grade nine word problems . the height of the rocket is the same as when it was fired . What altitude did the rocket reach is another question providing you know the altitude of the launch pad . Seeing that the rocket was fired and not launched it may have burned on the pad so the height of the rocket may be less depending on what material it was made of .........

Food for thought

I tried again to take up this challenged and began designing the rocket using a model rocket simulation program.

I gave the rocket a 13mm body tube. there are no commercially available motors to exactly equal the 2.5 N of impulse and definitely none that have a burn time of 8 seconds. So I tried with one that had 2.36 and staged them (3 stages) to gain the 8 seconds required.

My first attempt sent the rocket 736 meters, but my weight was too light.

I added mass to the first and second stages to bring the rocket to 1kg and still maintain a stable flight. You know that center of gravity and center of pressure thing.

I have never seen a simulated rocket crash on the launch pad before.

Maybe one of the parameters of the challenge is this takes place on another planet

GA 4

Not the first though

December, 1957: While the Soviets are putting large satellites and live animals into space, the Americans are having serious problems. Here, the unmanned Vanguard rocket explodes on the launch pad, a disaster carried on live television!

hot - isn't?

A rocket of 1 Kg mass requires 9.81 Newtons to counter the force of gravity to enable its lifting. Any force below 9.81 newtons cannot lift the rocket above ground level. The rocket will only burn the fuel off without lifting.

easy! 262.72ft (80.077m) high at 8 seconds velocity at 65.68 feet/sec(20.01m/sec) and max height at 267.94 feet(81.66m) at 12.079 second /flight stall.

Where does the extra 0.077m at 8-seconds come from? Double inaccurate conversion, perhaps - or are you assuming that the rocket becomes only very slightly lighter during its ascent? But after that the deceleration is slightly less than 1/4 of the original acceleration - so the rocket will go slightly more than a further 20-m.

Perhaps this was not "easy!" enough?

So in four seconds (more than expected) after flame-out , it travels about 1.6m (far less than expected)? The math is in fact easy, but your answer seems to be relying on something other than math.

Do CR4 forget to give the solutions of the monthly challenges or after seeing the responses decides the answer is controversial/ wrong ?

No CR4 solutions for

Planetary mass (Sept 2009)

Apple dealers (Oct 2009)

TO prove the rules have exceptions answered Polar Ice (Nov 2009)

and then continued where it left:

Flat Mirrors (Dec 2009)

Lamps & Switches (Jan 2010)

(Did not check earlier to Sept 2009, but the trend might have continued to medieval ages)

All true, but it's not Feb 2nd quite yet.

If you wish to influence events, maybe you should address this question to Chris Leonard?

Didn't include this one (Feb 2) the others - latest is the Lamps and Switches (Jan 5th) We are surely past that date ?

"It ascends subjected to a force of 2.5 Newtons acting vertically" = The rocket is subjected to a net force of 2.5 Newtons enough to lift the roket.

Answer #2 & #7 deserved the credits.

I should be #2 and #28

It is possible to determine the height of the rocket when it runs out of fuel - 80 m.

What is its speed at that moment - 20 m/sec

What is the maximum height reached - 100 m.

I assume the figures are correct and that the rockets weight is zero after it has burned all the fuel. With 2.5N force, the rocket will rest on the ground until its weight is 0.25 kg approximately, and after 6 seconds it will start moving. First very soft, then ever more fiercely screaming into the sky. When the last millisecond is reached, acceleration is 2.5N*1000/(9.81N/8)=2038g, approximately. The function reaches a pole, infinite acceleration, when the last molecule of propellant is burned. Numerical integration for the height, please. Certainly, the thing will blow up due to friction heat in air.

Rocket has zero weight? 2000g acceleration? Better switch to relativistic equations before that rocket squeezes through a wormhole into an alternate reality.

I have been studying the equations again and can not get the square root of numbers zero or less to calculate a positive altitude.

The idea of using a motor that burns up its mass until it light enough to lift itself off the ground is intriguing, but how do you calculate the maximum altitude of a zero mass after the 8 second burn. How far can a zero mass go during the coast phase? How would you stop it?

I vote for more parameters. Like mass of propellant, diameter and height of rocket, quality of construction, how many fins and their shape, or at least how much drag we should calculate.

it is not possible to solve this problem - important informations are missing!

the rocket could only lift of if in the time of 8 seconds the mass of the rocket decreases to less than 0,254929053244482060642523185797393kg!

If this does not the rocket could not lift off and so all asked values are zero!

Assuming the weight of the expelled fuel is insignificant, the rocket will reach a speed of 20 m/s at burn-out. The height at that point will be 80 m. Maximum height will be 100 m.

The rocket would not be propelled into the airs. a no start

The rocket weighs 1.0 kg = 9.80665 newtons

You only have 2.5 newtons of thrust (force)

Harry Peterson

Here We can split the motion of the rocket in to two phases.

PhaseI=> with Fuel

PhaseII=> without fuel, till reaching the maximum height, (or velocity=0 m/s)

Now for PhaseI, initial velocity, u=0.

Time of this motion is given as ,t= 8 s

Force is given as, F= 2.5 N

Now mass of the rocket ( please ignore the mass of the fuel), m= 1 Kg.

So the acceleration due to the force of 2.5 N for the rocket is F/m=2.5 m/s2.

So for phaseI, distance travelled, S= 1/2 * a* t^2= 1/2*2.5*64=80 m

So, the height of the rocket when it runs out of fuel is 80 m

Now for this motion (motion @ phase-I) final velocity, v=initial velocity,u+ at

So v=0+2.5*8=20 m/s

So, speed of the rocket, when it runs out of fuel is 20 m/s.

Now please consider the mtion @ phase-II

Here, initial velocity, u=20 m/s

Final Velocity, V=0 m/s.

Acceleration for this motion,"a" is 9.81 m/s2

So if "S" is the distance travelled, then V^2-u^2= 2*a*S=> S= V^2-u^2/2*a

ie; 0^2-20^2/2*9.81 (-signs cancell each other, because 9.81 is also -ve, retardation)

Therefore S= 20.39 m

So the Maximum height reached is 80+20.39=100.39 m

Yes, but this is a trick question.

The force of gravity on a 1kg rocket is 9.81 N acting vertically downwards. If you have a thrust of 2.5 N acting vertically upwards, the rocket will not leave the ground. so the height attained is 0 m

Assume that the 2.5N is the net force on the rocket

Assume g=9.81m/s^2

F=ma

therefore a=F/m = 2.5N/1kg = 2.5m/s^2

altitude when it runs out of fuel

s=ut+0.5*at^2

u=0m/s

t=8s

s=0.5*2.5m/s^2*(8s)^2=80m

Speed at that moment

v^2=u^2+2*a*s

v=sqrt(2*2.5m/s^2*80m)=20m/s

additional altitude

s=v^2/2g

s=20m/s^2/2*9.81m/s^2=20.38m

Total altitude =80=20.38 = 100.38m

This has got to be a trick question, to trap the guys who are so eager to work the numbers they can't even stub their toe on the obvious. I a motor/airframe combination can't produce enough thrust to more than equal the total mass, any aerodynamics are meaningless: It aint' goin' noplace. There might be a little spike due to a discontinuity in the fuel, but the best you could hope for there is to just clear the rail.

And nobody mentioned grain configuration. Is it an end burner or portburner? This would make a huge difference in the thrust curve, since the nozzle geometry would change significantly as the burn progressed. In solids, from homemade bottle rockets to ground-rumbling prime movers, nozzle erosion throws everything out of kilter, though this can be compensated for somewhat in the portburners. But to properly configure your grain, you'll have to have a pretty good handle on burn rate (generally .1in/sec.), propellant's thermal conductivity, thermal effects on sensitivity and rate, opacification (if any), thermal softening of the binder material (if any), and the list goes on.

This latter item mentioned is critical. If the srength of the grain, especially in shear, degrades too much, a chunk of propellant (AKA "sliver"), even small enough to easily pass through the throat, can break loose and either bounce around and further damage the grain, or shoot straight on out the back. Upon transiting the throat, it will usually cause a pressure spike sufficient to put you on the wrong side of the pressure/temperature curve & triggering autoignition (that's spelled B-O-O-M) and spoiling your fun. I've been in some of the control rooms at Aerojet & other facilities, and couldn't help but notice that in most of the older ones, huge, long cracks traveled through many of the walls. These buildings were of extremely heavy consruction, with walls 12 feet thick on the sides facing AWAY from the test stands, and over 20 on the front side. I asked a relative "old timer" about the cracks & whether they came primarily from solids. He said nearly all of them.

They used to test some real monsters out there, and windows would rattle for miles around. I could always tell if the test motor was solid or liquid by the duration and cutoff. A liquid can be shut down almost instantaneously; the sound (music?) would just stop. The sound of a big solid would kind of roll off as the last few remnants of fuel burned off. And once you've lit a solid, you're comitted. It'll turn off when it burns out. So for an all-up test, you'd have say, 2 1/2 minutes burn time, and that was it. If you didn't have all the data you wanted with that burn, you had to bolt a fresh motor in the stand, wire it up (you wouldn't believe all the sensors, although for solids it was considerably less involved in that regard), and shoot another test.

The LOX tanks and those for the various fuels (LH2, Kerosene etc) were in most cases much larger than those the engines had to actually fly. Frequenly, the opposite would occur. They'd get all the data they wanted from that particular set of parameters, or such and such an addition or modification, in maybe 45 seconds burn time and could just shut it down at will.

Trust me, I'm going somewhere with this. But it's 4:30 am & a guy's gotta' sleep sometime. I'll post the rest as soon as I've got my wits about me & have a little time.

Hi Guest; see post #7 or thereabouts.

CR4 may be repetitive - but it is frigging thorough

????

This month's Challenge Question:

You fire a 1.0 kg model rocket vertically from the ground. It ascends subjected to a force of 2.5 Newtons acting vertically. It travels for eight seconds before running out of fuel. After this, it continues upward until its velocity is zero and then it falls down to the ground. It is possible to determine the height of the rocket when it runs out of fuel? What is its speed at that moment? What is the maximum height reached?

I don't think you can solve this one unless you know the final mass of the rocket, or the velocity at which fuel is ejected, or the mass of fuel. Acceleration is Force (2.5N) / mass, but mass decreases as fuel is discharged from the rocket. Therefore, acceleration will increase at a rate inversely proportional to the rate of change of mass. Without knowledge of this, it looks to me as if it is impossible to know the velocity at t = 8s.

It may just do this http://www.youtube.com/watch?v=7Z-3fjg4dYY

"The Answer will be posted right here on CR4 on February 2nd. Can't wait---"

February 6, 2010

from all the posts, there still seems to be an interest of where this is leading us.