# How Much to Fill Bottle Rocket?

*Most recent answer: 05/02/2011*

- Kennedy (age 12)

Bethalto, Illinois United States

The energy to drive the rocket comes from the compressed air. (The water doen't compress enough to store much energy.) So you need pretty much of the rocket to be left empty. The forward momentum of the rocket has to just match the backward momentum of the material ejected. Air has so little mass that it can't carry much momentum backward. So you need the heavier water for that. So a starting guess would be to use 1/2 the volume each for air and water, and adjust around that by experiment.

I could get a little more precise, still not trying to do an exact solution. Call the fraction of the volume used for the water x, with (1-x) the fraction for air. The energy stored (at the maximum pressure you can use) will be proportional to (1-x). The mass ejected will be proportional to x. The energy that mass gets if it's ejected at speed v is proportional to xv

^{2}. So v

^{2}is proportional to (1-x)/x. So v goes as sqrt((1-x)/x)). The ejected momentum goes as mv, or sqrt((1-x)x)). That make it sound like our guess that the best x=0.5 is right (since it makes x(1-x) as big as possible), but we haven't allowed for the gradual change in speed as the rocket accelerates, etc., so that's just a starting point for your experiments.

Mike W.

*(published on 05/02/2011)*

## Follow-Up #1: bottle rocket calculation

- Lucy (age 12)

Canada

To do a more accurate calculation, you might do something like this.

1. Call the mass of the bottle m, and its volume V. We want to calculate x, the fraction of the bottle's volume to fill with water. The density ρ of the water is very close to 1.0 gm/cm

^{3}. So the initial mass is m+ρVx. Call the initial pressure p.

2. You have to figure out how much momentum each little increment of expelled water gives to the remaining rocket, whose mass is the rocket mass plus the remaining water mass. The change in momentum of the remaining rocket is equal to the speed s of the ejected mass in the frame of the rocket. I don't quite know how to calculate s because I'm not sure how it depends on the pressure. The pressure drops as the empty volume grows, approximately inversely proportional to volume. (For specialists- I'm using isothermal, although adiabatic may be closer.)

3. The speed change of the rocket is s*(ejected mass) / (remaining mass)

4. Since the pressure and the remaining mass keep changing, you have to do an integral over little increments of ejected mass to get how the rocket speed changes as the mass gets ejected. You also need to use the changing s to calculate how much mass is ejected per time, so that you can convert that to distance traveled.

So that's the outline. Even this calculation involves some approximations, such as neglecting air friction. You see why I recommended just trying a few different fill levels around the 1/2 filled value. That saves doing a tough calculation and avoids any approximations or forgotten influences in the calculation.

Mike W.

p.s. I tried playing around a little with doing that integral, still using some approximations. It looks like in the approximate calculations for a range of guesses about the weight of the bottle somewhere around 30% filled gives about as good a result as you can get.

*(published on 05/08/2011)*

## Follow-Up #2: bottle rocket plans

- Lucy (age 12)

Quebec

We've had a lot of discussion about how much water to use. I did a very crude calculation that gave 1/2 full as a suggestion. Looking around, I see that many people with experience recommend about 1/3 full. The main factor left out of my calculation was allowance for the first water expelled having to lift not just the bottle but also the water that hasn't yet been expelled. Including that would reduce the recommended level below the 1/2-fill found in the crude calculation. So 1/3 full sounds like a good starting point.

The only reason I can think to use several bottles is to reduce the ratio of air friction to rocket drive. I bet that isn't very effective at all, since the air friction approximately doubles as you double the number of bottles. I'd just use a single big bottle and put a lot of care into making good fins, etc. and into testing everything carefully. This is just a guess, but I bet that works better than fussing with a bunch of bottles.

If for some reason you have to use more than one bottle, duct tape would be ok.

Mike W.

*(published on 05/09/2011)*

## Follow-Up #3: bottle rocket plans

- cherri (age 14)

kentwood,mi

If you want the wings to be symmetrically arranged (balanced) and to provide stability against wobbles in all directions, you need at least 3. Wings also give some drag, so I would keep it to either 3 or 4. You could of course use some tape, but if you can get things to hold well enough with glue, that might add less weight. I don't know how to figure the best slope for the wings, so you might just look at what other successful bottle-rocketeers have used.

Mike W.

*(published on 05/24/2011)*

## Follow-Up #4: big or little bottle rocket

- Charl (age 15)

Boston

Mike W.

*(published on 06/06/2011)*

## Follow-Up #5: bottle rocket fins

- shujaat (age 20)

Aurangabad, Maharashtra, India.

Mike W.

*(published on 07/03/2011)*

## Follow-Up #6: filling a bottle rocket

- Savannah (age 13)

Andover mn usa

Again a reminder- for quick answers, be sure to use our search engine. Often, as here, we've already answered your question.

Mike W.

*(published on 09/17/2011)*

## Follow-Up #7: filling bottle rocket

- Shaniqua

Mike W.

LeeH

*(published on 09/28/2011)*

## Follow-Up #8: why 1/3 filled bottle rocket

- Samantha (age 14)

Illinois

Hi Samantha- Great question! Unfortunately I can't give a great answer. Like we wrote before, there's a balance between too much air vs. too much water. If the whole bottle was filled with water, even under pretty high pressure it just wouldn't compress enough to store much energy. If it was all filled with compressed air, you could store a lot of energy but there wouldn't be enough mass ejected from the nozzle to impart much momentum to the bottle. The 1/3 is just a rough estimate based on balancing these factors.

Mike W.

*(published on 05/22/2019)*