Notmadrocketscientist rocket launch

[totalitor]

Hi!

 

I have been designing and building small rocket engines few years and now I have a rocket almost ready.

It is passive fin stabilized. Wet mass is 3 kg. It consumes propellants about 35 g/s (average), burn time is 10 seconds.

Thrust curve is 1 second intervals in newtons: 44-41-39-37-33-31-29-27-24-22

Yes, I have problems keeping thrust steady.

I have tested that fins do work at speed of 10 km/h. Look video here: https://www.youtube.com/watch?v=6nSyWYE1VlM

Now I need to build launch stand/rail/tower.

Can anyone calculate lenght of launch rail to achieve that 10 km/h speed?

And what is estimated apogee?

 

[caecilliusinhorto]

Have you tried modelling the rocket in some form of simulation software? OpenRocket is quite good and is completely free, it will give you altitude, velocity and lots of other values for different engines and can plot them.
 

[sevenperforce]

17 hours ago, totalitor said:

It is passive fin stabilized. Wet mass is 3 kg. It consumes propellants about 35 g/s (average), burn time is 10 seconds.

Thrust curve is 1 second intervals in newtons: 44-41-39-37-33-31-29-27-24-22

Yes, I have problems keeping thrust steady.

It doesn’t look like you’re having trouble keeping thrust steady; it looks like you’re using propellant blow-down, and so the thrust drops off more or less linearly. Slope of the thrust curve appears to be around -2 Newtons per second.

17 hours ago, totalitor said:

I have tested that fins do work at speed of 10 km/h.

Can anyone calculate lenght of launch rail to achieve that 10 km/h speed?

Obligatory: please be safe. Don’t be anywhere close to the thing when you light it. Obey all local laws and don’t assume you know what the laws are. Make sure there is absolutely 0% chance that anything flammable is downrange.

That said, this is a fairly simple kinematics problem. If your engine burns an average of 35 g/s then your mass will be (at 1-second intervals, in kg): 3.0-2.965-2.93-2.895-2.86-2.825-2.79-2.755-2.72-2.685-2.65.

Newton’s First Law helpfully reminds us that F=ma, so a=F/m. Liftoff acceleration will be 14.67 m/s^2 and burnout acceleration will be 8.3 m/s^2. However, gravity will be tugging the rocket back down toward terra firma the whole time at a cruel 9.81 m/s^2, so the actual liftoff acceleration will be 4.86 m/s^2. 10 km/hr is 2.78 m/s, so your rocket will exceed stabilization speed 0.57 seconds after launch.

Assuming constant thrust for simplicity and starting from a standstill, x = a(t^2)/2, so your rocket will cover 0.79 meters before reaching stabilization speed. However, this doesn’t account for factors like the drag of the stabilization system or the drag with the launch rail interface or startup thrust gradients or any other number of variables. So to be safe you should probably do a 200% safety factor and make your launch rail about 2.4 meters high.

[sevenperforce]

17 hours ago, totalitor said:

And what is estimated apogee?

By the time you get to burnout, thrust will have dropped below one gee, meaning that it will start to lose velocity due to gravity drag before it completely burns out. But if you calculate the specific impulse and then the delta-v, you can deal with a lot of this easily.

I have absolutely no idea what your drag coefficient looks like. Honestly, I think you would do better with fewer, larger fins, but that’s completely heuristic advice so your mileage may vary. Without any concept of your velocity-specific drag coefficient I can’t hope to speculate about your rocket’s actual performance, but what I can do is put an upper limit on your apogee, based on what you’ve said.

Specific impulse is the inverse of fuel-flow-specific thrust and can be calculated as the thrust divided by the propellant mass flow rate. Fortunately, you’ve provided both. Unfortunately, you don’t have constant thrust. I could assume that you’re using a pressure-fed propellant-blow-down design, resulting in a linear mass flow drop rate, but since I’m just going for the upper limit, I don’t need to. Your max thrust is 44 Newtons, which divided by 0.035 kg/s gives us a specific impulse of 1,257 m/s or 128 seconds.

What’s your propellant combination and chamber pressure? Not that it matters for calculating the apogee; I’m just curious. You’re getting choked flow, right?

Anyway, if you apply the Tsiolkovsky rocket equation, you get a total dV of 156 m/s. However, you’re dealing with gravity drag. Ten seconds of gravity drag robs you of 98.1 m/s, so your actual effective dV is going to be 58 m/s. So that’s your theoretical maximum burnout velocity.

Given the thrust curve, mass gradient, and burnout velocity, plus the generous and inaccurate assumption of negligible aerodynamic drag, anyone equipped with first-semester physics can calculate an apogee.

[caecilliusinhorto]

The propellant combination is ethanol and oxygen gas which is explained in one of the videos.

[totalitor]

Sevenperforce: I knew it was you who make calculations. Thank you.