How far will a toy car travel with a small rocket motor

[marciokoko]

I want to strap an estes b4-4 rocket to a toy car and calculate how far it would travel.

What I know:

Mass of car: 300 grams
Rocket nominal force: 12.8 Newtons

I think this means the rocket would accelerate my car for a=F/m = 12.8 kg * m/s^2 / 0.300 Kg which means 12.8 m/s^2/ 0.3 = 42.67 m/s^2.

Where do I go from here, assummig negligible friction forces?

 

[DrGreg]

Assuming negligible friction forces it will travel an infinite distance. Bad assumption!

 

[sophiecentaur]

Rocket nominal force: 12.8 Newtons

How long is the rocket burn? Force times time is Impulse, which is the change in momentum.
That will allow you to work out the speed of the car at the end of the burn.
You can then work out how far the car will travel in a given length of time.
If you have ever used the SUVAT equations, you could also easily work out how far the car will travel as it is accelerating.

 

[marciokoko]

Total impulse it says 4.3 Ns and burn time 1s.

So should I do 42.67 m/s^2 * 1s = 42.67 m/s? And then use s = 1/2(u+v)t = 1/2(0+42.67m/s)1s = 1/2*42.67 = 21m?

or

do I just do something straight with 4.3 Ns? I am still getting my head around this one.

DrGreg
Ok I must assume friction forces...what would be the simplest calculation, just a downwards gravity based vector due to mass?

 

[sophiecentaur]

do I just do something straight with 4.3 Ns?

yes. The unit Ns tells you that the same result is achieved with 1n for 10s or 10N for 1s. (Like Amp Hours in a battery)
So your momentum will be Mass X velocity and mass is 0.3kg so you can work out the final speed.
4.3 = 0.3v
v= 14.33'm/s
But my point about a "given length of time" is relevant because if this is about a real experiment, the track may have to be ridiculously long for a low friction car to slow to zero. Have you enough room (starting at around 30mph)?
PS When you write out an equation, it's not usual to include the units. Just put them in the answer. Very confusing inside an equation.

 

[marciokoko]

OK wait I got that's it's final velocity will be 42m/s whereas you got 14.3 m/s.

With 42m/s using a suvat I get 21m displacement. With your final velocity it would be about 1/4 of that, so around 5m displacement.

Where am I wrong?

 

[sophiecentaur]

OK wait I got that's it's final velocity will be 42m/s whereas you got 14.3 m/s.

With 42m/s using a suvat I get 21m displacement. With your final velocity it would be about 1/4 of that, so around 5m displacement.

Where am I wrong?

Not 'wrong' - just confused by what they have told you.
If they specify the impulse as 4.3Ns then that is the safest thing to work with. They will have measured it for several motors and it's a reliable method (standard way to measure these things, I believe). If they say that the Force it produces is 12.8N, they may mean that is the maximum force during the burn. If the burn is 1s then their 12.8N figure can't be right for the whole burn time or the impulse would be 12.8Ns and it can't be both values, can it? I suspect the 1s figure, actually.
Stick with the 4.3Ns figure because it's the easiest to measure and, short of a typo on the spec sheet, it's likely to be more reliable as it doesn't involve time.
Edit: It's worth while imagining what goes on in the solid fuel as it burns. It will start of as a fizz, then the burn will work its way over the total area of the cylinder, then some sort of cone shape will develop (max power here, probably) and then gradually fade out. If that's to take up the whole 1s of time, it's not surprising that the average Force is only applied, effectively, for around 1/3 of the burn time. This could easily account for the 12.8:4.3 ratio of forces that the two calculations used.

 

[marciokoko]

Yeah, I had considered that as well, that the max thrust, stated precisely as Max, would just be a brief peak compared to the get-up and come-down parts of the burn.

OK so using s = 1/2(u+v)t and your final velocity of 14.33m/s I get:

1/2(0+14.33)1 = 7.15m

That's too little. These things go 750' up with a 1 ounce rocket. According to my calcs, a 1 ounce estes rocket would weigh 28g which would put its final velocity at 153 m/s vs the 14.3 m/s of my 300g rocket. Those velocities plugged into 1/2*(u+v)*t would put their respective displacements at 7.15m and 76m. Like I said, 7.15m seems like a very short distance to travel, but I haven't done it. But I am sure the estes rockets go higher than 76m, 251ft.

So I need more tweaking.

 

[sophiecentaur]

Are we talking rockets now?

OK so using s = 1/2(u+v)t and your final velocity of 14.33m/s I get:

1/2(0+14.33)1 = 7.15m

This is not the equation to use, for a start. You are assuming that the rocket flight is only 1s, which obviously makes it all a bit dodgy. To find the height achieved, you can use the 'other' method by equating the KE at the start (mv²/2) with the PE at the top (mgh). This doesn't require you to find the flight time.
h=v²/2g
= 42m approx
Go over all this again and you may get more sense from it.
When using the suvat (and any other) equations, it is essential that you can guarantee that you know the values of what you are putting in and make no assumptions. There is usually a way through to an answer but you have to choose the right one.

 

[marciokoko]

No, I am still talking cars in a straight line. But I am just comparing a 7m displacement sideways vs a 209m displacement upwards. Its a big difference so it seems its too little a horizontal distance of 7m if vertically (albeit lighter rocket vs car) its supposed to climb 209m.

What I am still trying to figure out is how long the car should travel.

 

[jbriggs444]

What I am still trying to figure out is how long the car should travel.

See the first reply in this thread.

 

[marciokoko]

The negligible friction forces? Ok so could you explain to me how to calculate the distance it will travel including friction? Because if anything, those will make the calculated displacement of 7.15m even smaller, which I am pretty sure won't be the case when I light it. A small wooden car with the b44 motor strapped to it will surely travel more than 7.15m in a straight line.

 

[sophiecentaur]

But I am just comparing a 7m displacement sideways

You are in a muddle with this, you know. If it is traveling at 14.3m/s how can it only get 7m, even in the first second of its travel? Which suvat equation suits that question? (The easiest one)
Also, if there is no friction, how will the car ever stop on a horizontal track? Think a bit more logically and don't just take random equations to give you random answers.
Just how do you get your 7m figure?

 

[marciokoko]

OK well in post #8 I showed how I got it using equation s = 1/2 * (u+v) * t but I guess the 1s is too little. I took it from the rocket datasheet which states Burn time as 1s. initial velocity I used zero and final velocity the 14.3m/s calculated from your post. Why is that not the right equation?

 

[sophiecentaur]

But what has the time it travels during burn time got to do with how high the quoted rocket reaches? Also, that formula assumes constant acceleration. This is why we use Impulse.
Can you please decide what it is you actually want to find out?

 

[marciokoko]

Well I've always wanted to just find out how far a car will travel if I strap that rocket motor to it.

 

[jbriggs444]

OK well in post #8 I showed how I got it using equation s = 1/2 * (u+v) * t but I guess the 1s is too little. I took it from the rocket datasheet which states Burn time as 1s. initial velocity I used zero and final velocity the 14.3m/s calculated from your post. Why is that not the right equation?

The car does not stop moving just because the rocket stops burning. Same as the rocket does not stop rising just because it stops burning.

 

[marciokoko]

I understand that, so what is the proper way to calculate how far my car will travel?

 

[jbriggs444]

I understand that, so what is the proper way to calculate how far my car will travel?

Come up with a model that describes the friction it will encounter as it travels.

 

[marciokoko]

A model? Like a vector drawing or something like that? I am not a physicist but I am guessing something like this:

 

[jbriggs444]

In physics, a "model" or "mathematical model" is a set of formulas or equations that describe a phenomenon. One simple model for friction is that it is proportional to the perpendicular force between an object and a surface. The perpendicular force is referred to as the "normal" force. The constant of proportionality is the coefficient of kinetic friction. The greek letter mu ($\mu$) is conventionally used to denote the coefficient of friction. The coefficient of friction needs to be measured.

$$ F_{friction} = \mu F_{normal} $$

 

[sophiecentaur]

I understand that, so what is the proper way to calculate how far my car will travel?

Give us the scenario in full and we have a chance of giving you the answer.
I am disappointed that you appeared, earlier, to be taking some of the main principles on board but now you are back with the same vague question.
Suvat does not work for the burn time because suvat describes motion under CONSTANT acceleration.

 

[Nugatory]

A model? Like a vector drawing or something like that?

No, in this context a "model" is some reasonably accurate quantitative description of the friction forces. These are the forces that will slow down and eventually stop the car, so you must have some idea what they are to calculate how far the car goes before it stops.

 

[AlphaLearner]

I say, as long as the force exerted by the rocket motor is greater than that of frictional force, the car won't stop. It keeps moving. So the time period taken for car to stop in that case is 'Infinite'. Once the force gets lesser than that of friction force, the car can stop at some point and yes, its time period can be calculated in this case.

 

[marciokoko]

Well I didnt think i was asking such a complex question. Ok I have a car to which I have strapped a b4-4 estes rocket motor on top. I plan to set the car on the street, light the motor and let it go as far as it will go on its own until the motor burns out.

 

[sophiecentaur]

Well I didnt think i was asking such a complex question. Ok I have a car to which I have strapped a b4-4 estes rocket motor on top. I plan to set the car on the street, light the motor and let it go as far as it will go on its own until the motor burns out.

So, as I said, you have an impulse (see the spec) and that tells you the speed at the end of the burn. That is not rocket science (errr um). The question is how far will the car go after the burn is finished - and that is a lot further than the distance during the burn (unless it is a ploughed field).
The question is as "complex" as you want to make it but there is a bare minimum of conditions that you have to be prepared to specify.
It's not surprising that people start their investigations with ballistic behaviour. It's so much simpler than cars on tracks. Take it a step at a time.

 

[jack action]

You have 2 resistances: rolling resistance $F_R$ and aerodynamic drag $F_D$. Aerodynamic drag should be the greatest by far.To get the acceleration, it is simply:
$$\sum F = ma$$
$$F - F_R - F_D = ma$$
$$a = \frac{F - F_R - F_D}{m}$$
Where $F$ is the force produced by your rocket.

During the travel time, these forces will vary, so the acceleration will also vary. For the rocket, I'm not an expert on these engines so I can only guess on the force output versus time $F(t)$ equation to define it. Maybe other can chime in. But let's imagine that it is constant for the full 1 s.

The rolling resistance is mathematically similar to friction, so $F_R = C_{rr} F_N$, where $C_{rr}$ is the coefficient of rolling resistance and $F_N$ is the normal force (i.e. the vehicle weight $mg$ in our case).

The aerodynamic drag is $F_D = \frac{1}{2}\rho C_D A_f v^2$ where $\rho$ is the air density (1.23 kg/m³), $C_D$ is the drag coefficient and $A_f$ is the frontal area of the vehicle and v is the speed of the vehicle. As you can see, as the vehicle increase in speed, the drag increases, thus your acceleration will get smaller as it travels.

To calculate the velocity after 1 s of travel (when the rocket pushes the vehicle):
$$dv = adt$$
$$dv = \frac{F - C_{rr} F_N - \frac{1}{2}\rho C_D A_f v^2}{m}dt$$
Which can be rewritten as:
$$\frac{dv}{\frac{F}{m} - C_{rr} g - \frac{\rho C_D A_f}{2m}v^2} = dt$$
Simplified as:
$$-\frac{1}{B}\frac{dv}{v^2 - C^2} = dt$$
or
$$t = -\frac{1}{B}\int_0^{v_f}{\frac{dv}{v^2 - C^2}}$$
Where:
$A = \frac{F}{m} - C_{rr} g$;
$B = \frac{\rho C_D A_f}{2m}$;
$C = \sqrt{\frac{A}{B}}$.
The solution of the previous integral is:
$$t = -\frac{1}{2CB}\ln\left|\frac{v_f-C}{v_f+C}\right|$$
You can find $v_f$ for $t=1$ by trial and error. Check that $a(v_f) > 0$. If it is not, then calculate the speed $v_f$ that will give you $a = 0$ and find times $t'$ that it will take to reach that speed with the previous equation. Then the vehicle will be going at constant speed $v_f$. $v_f$ when $a=0$ is known as the terminal velocity of your vehicle.

After you found $v_f$ (and $t'$ if that was the case), you can find the distance traveled with:
$$dx = \frac{v}{a}dv$$
$$dx = \frac{v}{\frac{F}{m} - C_{rr} g - \frac{\rho C_D A_f}{2m}v^2}dv$$
Simplified as:
$$dx = \frac{v}{-Bv^2+ A}dv$$
or
$$x = \int_{v_0}^{v_f}{\frac{v}{-Bv^2+ A}dv}$$
The solution to this integral (see 17 here) is:
$$x = \frac{\ln\left|-Bv_f^2+A\right|}{-2B} - \frac{\ln\left|-Bv_0^2+A\right|}{-2B}$$
Where $v_f$ was found previously and $v_0$ would be the initial velocity, i.e. 0.

If there was a $t'$ found previously, the vehicle will travel from $t'$ to $t$ at constant speed $v_f$, thus you must add $v_f (t- t')$ to $x$ to find the total distance traveled during the full time $t$.

The rocket stops

Now your vehicle is going at speed $v_f$ and has traveled a distance $x$ after time $t=1\ s$ (or whatever other time you want). The rocket stops pushing, thus $F=0$ at this point and $A$ is modified to $A' = - C_{rr} g$. What we want to know is how far will it go before it stops by itself (due to drag and rolling resistance).

We know the initial speed $v_f$ at this point and we know final speed will be 0 when it stops. Thus we can use the same equation as before, but with the new values:
$$x' = \frac{\ln\left|-B(0)^2+A'\right|}{-2B} - \frac{\ln\left|-Bv_f^2+A'\right|}{-2B}$$
And the total distance $d$ traveled will be:
$$d = x+x'$$

Test
With the following data:
$m = 0.3\ kg$;
$F = 10\ N$;
$C_{RR} = 0.01$;
$C_D = 0.4$;
$A_f = 0.1\ m \times 0.1\ m$.

I get for $t=1\ s$:
$v_f = 30.5\ m/s$;
$x = 15.9\ m$;
$d = 282.2\ m$.

If there is a discrepancy between those numbers and my equations, there is probably an error in my equations when I transcribed them.

 

[Nugatory]

Well I didnt think i was asking such a complex question. Ok I have a car to which I have strapped a b4-4 estes rocket motor on top. I plan to set the car on the street, light the motor and let it go as far as it will go on its own until the motor burns out.

Have you noticed that this is NOT the question you asked in your original post? In your original post you asked "how far it will travel", and because it will keep on moving after the motor burns out that will be considerably farther than just "until the motor burns out". And you're right that neither of these are such complex questions, but even the answers still require more data than you have provided:

If you meant to ask "How far will it go?", we need an estimate of the frictional forces acting on the car, because that's what stops it.

If you meant to ask "how far will it go until the motor burns out?" we need to know how the acceleration varies over time. You might try assuming a constant acceleration consistent with the total impulse - that's easy to calculate and probably won't be off by more than a factor of two or thereabouts. Note that even if the thrust from the motor were constant the acceleration would not be be constant because the mass being accelerated changes as the motor loses mass during the burn; therefore constant acceleration has to be an approximation at best.

And one other stray question: Why are you using a B4-4 motor instead of a B4-0? It doesn't sound like you're planning to deploy some sort of braking device, so why are you planning on that little explosion out the front end four seconds after the burn has ended?

 

[marciokoko]

OK I'm sorry for my ignorance, I don't see them as different questions. If I put the car at the beginning of the street and light it, how many meters from the starting point will the car end up at?

I didn't remember about the parachute explosion at the end of the burn but I'm using it because that's all I have :-)

OK so the force of the rocket will push it forward and the fictional forces will stop it. How do I calculate those forces in order to know where it will stop?

Have you noticed that this is NOT the question you asked in your original post? In your original post you asked "how far it will travel", and because it will keep on moving after the motor burns out that will be considerably farther than just "until the motor burns out". And you're right that neither of these are such complex questions, but even the answers still require more data than you have provided:

If you meant to ask "How far will it go?", we need an estimate of the frictional forces acting on the car, because that's what stops it.

If you meant to ask "how far will it go until the motor burns out?" we need to know how the acceleration varies over time. You might try assuming a constant acceleration consistent with the total impulse - that's easy to calculate and probably won't be off by more than a factor of two or thereabouts. Note that even if the thrust from the motor were constant the acceleration would not be be constant because the mass being accelerated changes as the motor loses mass during the burn; therefore constant acceleration has to be an approximation at best.

And one other stray question: Why are you using a B4-4 motor instead of a B4-0? It doesn't sound like you're planning to deploy some sort of braking device, so why are you planning on that little explosion out the front end four seconds after the burn has ended?

Ok

Have you noticed that this is NOT the question you asked in your original post? In your original post you asked "how far it will travel", and because it will keep on moving after the motor burns out that will be considerably farther than just "until the motor burns out". And you're right that neither of these are such complex questions, but even the answers still require more data than you have provided:

If you meant to ask "How far will it go?", we need an estimate of the frictional forces acting on the car, because that's what stops it.

If you meant to ask "how far will it go until the motor burns out?" we need to know how the acceleration varies over time. You might try assuming a constant acceleration consistent with the total impulse - that's easy to calculate and probably won't be off by more than a factor of two or thereabouts. Note that even if the thrust from the motor were constant the acceleration would not be be constant because the mass being accelerated changes as the motor loses mass during the burn; therefore constant acceleration has to be an approximation at best.

And one other stray question: Why are you using a B4-4 motor instead of a B4-0? It doesn't sound like you're planning to deploy some sort of braking device, so why are you planning on that little explosion out the front end four seconds after the burn has ended?

 

[Nugatory]

OK I'm sorry for my ignorance, I don't see them as different questions.

Go back to your post #8, where you wondering why you calculated 251 feet when you were expecting 750+ feet. 251 feet was the answer to the question "how far does the rocket go before the motor burns out" and 750 feet or more was the answer to the question "how far does the rocket go?" That's a big difference.

 

[Nugatory]

How do I calculate those forces in order to know where it will stop?

You don't calculate them. They're inputs into the calculation so you have to know them up front, just like you know up front that the car has a mass of 300 grams. That's why people are telling you that you need a model for the friction, or otherwise there isn't enough data to solve the problem.

If this were my experiment I'd think I'd go ahead and try it and see how far the car goes, use that data to tell me what the frictional forces are. It will probably take a few tries to get good data, but then again you'll probably need a few tries anyway to get dynamic stability and straight-line tracking from your car.

 

[jack action]

If this were my experiment I'd think I'd go ahead and try it and see how far the car goes, use that data to tell me what the frictional forces are. It will probably take a few tries to get good data, but then again you'll probably need a few tries anyway to get dynamic stability and straight-line tracking from your car.

So if someone asked you «how do I calculate the maximum weight a rope can hold?», you would answer «add weights to the rope and see when it breaks.»

Where's the physics in that answer? The fun is about predicting the outcome, not observing it.

The friction forces are well known: Rolling resistance and aerodynamic drag. There are enough values available online to make good estimates of these forces.

By the way, I thought the question was obvious from the original OP.

 

[marciokoko]

Nugatory:
That was not my original question. That was just a comparison to what the motor could do for the rocket. My original question could only be, by definition, in my original post, #1. In any event, I think I understand your point, there is a distance the object will travel while the motor is burning and there is an additional distance traveled after the motor has burnt out. Just to clear things up, I need both distances, in other words, the distance from where the car starts, to where it stops.

Yes I could try it but I ran out of igniter clips so I need to get some more. In the meantime I am trying to get an estimate.
Yes I understand I need a model for the friction but it's painfully clear that I don't know how to come up with such a model which is why I am asking for help with that since post #19 when jbriggs suggested it and I made my little drawing.

So I know the car will have a force pushing it forward, given by the 12.8N stated by the rocket datasheet?

The force acting against it is friction as defined by u * F(normal).

jackAction thanks for your post.
I would use a stagecoach reference 0.0385 for Crr
and 1.28 flat plate perpendicular to flow and 16 cm2 for my Af.

Great explanation on the integrals. You actually got me interested in them again because your explanation demonstrates how an integral is used to calculate the problem. Of course you simplified and solved them for me which was a great help btw. I was able to follow most of the derivations though. Now what do you mean by "I can find vf at t=1s by trial and error"?

Ok now with vf I can calculate x traveled and then add the next part which is the distance traveled now that F = 0.

I love it!

 

[arydberg]

With no friction the car will go on forever. That is why you need to know the friction. I would assume it is very low so your car would go a long distance. To measure the friction place the car on a flat board and slowly tip the board until the car begins to move. Measure the angle. This tells you the friction but does not include air resistance you need to be moving through the air at the terminal velocity.

 

[Nugatory]

So if someone asked you «how do I calculate the maximum weight a rope can hold?», you would answer «add weights to the rope and see when it breaks.»

Of course not. I'd tell them to start with the known tensile strength of the rope material and calculate from there and the diameter and construction of the rope (and of course adding an appropriate safety factor). But where did that known tensile strength come from? Or equivalently, what would we do if the rope was made of a new material with hitherto unmeasured properties?

The fun is about predicting the outcome, not observing it.

Well, speaking as someone who actually has attached Estes rocket motors to rolling toy cars... The observation is no less fun, and getting first dynamic stability and then repeatable results is quite challenging.