The Blitzkreig (WARNING LARGE PICS!!)

Threads related to water balloon launchers.
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Lightbulb41
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The Blitzkreig (WARNING LARGE PICS!!)

Post by Lightbulb41 » Sun Jun 04, 2006 1:17 am

I have a design for a new gun... I call it the Blitzkreig or the MOAC (Mother of all Cannons) It has a range of about 1/2 mile given 4 pound balloon 75 psi tank at a 45 degree launching angle. With a 12 foot barrell Here is the basic concept
Image


I found a tool that would help me calculate the size of the barrel I needed given ceartain information and Give me the maximum height of the projectile in meters

Using the range formula 4*(Max height)= Range I found that the max is about 1000 meters. To convert Meters to Feet I multiplied the 1200 * 3 to give me feet and Voila a 3000 foot range gun at 75 Psi... The problem is that this gun is so massive that It can only be mounted on a frame. Here is a Graph of what Math says the range could be...

Image

Remember that what it says for max height you multiply by 4 to get the range in meters and then multiply it by 3 to get feet. Then you take a bit off to account for air resistance.
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SSCBen
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Post by SSCBen » Sun Jun 04, 2006 1:36 am

If this tool is telling you that you have a 1/2 mile range and does not factor in air resistance, then you need to take off much more than a little to account for air resistance. Range also is not directly related to maximum height with the angle as a variable, and you can not find range with a formula based upon maximum height. Of course, with a fixed angle, you can make a formula to find that, but I'd feel safer using real math. To summarize: don't expect a half mile range or even a quarter mile range.

If this spreadsheet can take some drag coefficient, definitely try that. You might also have to determine the drag between the barrel and the water balloon, but that would probably be beyond the abilities of the spreadsheet because I have not seen a spreadsheet or program that does that yet (given that working with that requires differential equation knowledge).

Other than that, this appears to be a very straightforward water balloon launcher and I don't feel that you'll have any trouble constructing it. Good luck. :)

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Silence
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Post by Silence » Sun Jun 04, 2006 1:55 am

I doubt you'll get such range, but we'll wait and see. No WBL can have that poor a range...

What is the practicality of such range? I mean, if you can't even see your target, let alone aim properly and let alone wind and other factors that will throw off your aim, this seems a little pointless.

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Lightbulb41
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Post by Lightbulb41 » Sun Jun 04, 2006 2:13 am

I Plan to have the angle fired upon at a constant said so in beginning of the model. I also took off about 10% of the distance due to air resistance.


Also I plan to have this on a mounted turret... Degree markings and everything. All I need is a Laser pointer and a scope and I'll be able to hit a target with extreme presision due to speed. wind won't be a problem considering I'm firing a 4.4 pound balloon
You can get more with a kind word and a gun than just a kind word.

I trade mobility for firepower. Accuracy for strengh. And Stealth for Intimidation.

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Lightbulb41
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Post by Lightbulb41 » Sun Jun 04, 2006 2:54 am

I checked most of the calculations manually. It was difficult converting PSI to pounds force to Joules but the Difference is only slight. I think shaving 10% off the final answer is reasonable. It will be interesting when I have time to actually build this gun (after finals) and test it in battle (Massive water ballon fight with multipe launchers shelling different camps no hand to hand combat :( )
You can get more with a kind word and a gun than just a kind word.

I trade mobility for firepower. Accuracy for strengh. And Stealth for Intimidation.

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Hunt_and_Annoy
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Post by Hunt_and_Annoy » Sun Jun 04, 2006 3:16 am

I would prefer 2 or three compact WBL's to one gigantic impractical one any day...

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Drenchenator
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Post by Drenchenator » Sun Jun 04, 2006 1:40 pm

Good job on the design. But I would not trust using a spreadsheet to do the calculations. A half mile range is seemingly impossible and unusable, and I believe that the same goes to a 400 foot range too.

I have a few questions about the barrel length graph. I assume that it optimizes the barrel length for range. The height of projectile does not determine the range, so I do not believe that this graph would really be of any use. Also, some of the numbers that you entered seem odd. A 6 inch diameter barrel appears to be too large, and the fictional force experienced by the object in the barrel should be more the 4 newtons (1 newton = about .22 pounds of force). 4 N is the weight of 400 mL of water (400 g) which is not a lot (I would expect more). Did you use a force meter to measure this?

I am sorry if I appear negative, it is just that I prefer to do all of my own calculations. Don't give your hopes up, most of these launchers have good ranges (300+ feet) and you design seems sound. The only thing that I do not understand in it is the need to have a second ball valve after the check valve, the check valve would be all that you need.

EDIT:
I also took off about 10% of the distance due to air resistance.
Ten percent off for air resistance is incorrect. If anything, 50 to 70 percent off for air resistance is reasonable. I remember a graph in my physics book of a baseball as a projectile with and without air resistance. The graph with air resistance had over twice the range of the one without. And this was of a baseball, something that is basically a sphere.
Last edited by Drenchenator on Sun Jun 04, 2006 1:46 pm, edited 1 time in total.
The Drenchenator, also known as Lt. Col. Drench.

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Lightbulb41
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Post by Lightbulb41 » Sun Jun 04, 2006 3:09 pm

Agreed that the barrel force is not as accurate as I would like but. Remember as you fire the rate could go up or down. The thing is that I can always compensate for the lost range by increasing the pressure. The reason I have a second valve is because the compresser I'm using doesn't have a good checkvalve in it So I have to seal the pressure off for a quick time while I pull the other valve to release the balloon. The reason I didn't take off more for air resistance because I'm lazy right now. You are probably right that this has a much shorter range due to air resistance but the formula is correct for changing max projectile height to Range

Check it out here http://www.xs4all.nl/~mdgsoft/catapult/ballistics.html

Remember 4*(Max projectile height) = range

also remember Image

for knetic energy. and converting that and solving with the conversion PSI times area of the barrel to get pounds force multiply by 1.356 to get Joules and solve from there.
You can get more with a kind word and a gun than just a kind word.

I trade mobility for firepower. Accuracy for strengh. And Stealth for Intimidation.

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Silence
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Post by Silence » Sun Jun 04, 2006 3:12 pm

@ Lightbulb41: Umm, at the top of the page you linked to, there's bold text that says all the measurements assume there's no fluid air resistance. Sorry, but that probably means all the measurements are off.

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Lightbulb41
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Post by Lightbulb41 » Sun Jun 04, 2006 3:14 pm

Correct. That is because I haven't got around to calculating the air resistance quoutient into the equation yet (Google is confusing me)
You can get more with a kind word and a gun than just a kind word.

I trade mobility for firepower. Accuracy for strengh. And Stealth for Intimidation.

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Lightbulb41
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Post by Lightbulb41 » Sun Jun 04, 2006 3:24 pm

Found it . Heres the aproximate formula for air resistance I am not sure just yet how to calculate that in.
Image

cd= coefficient of friction,
a= frontal area of the object;

p=Greek letter rho, density of air, which we calculate below;

v=speed of the object.
You can get more with a kind word and a gun than just a kind word.

I trade mobility for firepower. Accuracy for strengh. And Stealth for Intimidation.

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SSCBen
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Post by SSCBen » Sun Jun 04, 2006 5:35 pm

also remember Image

for knetic energy. and converting that and solving with the conversion PSI times area of the barrel to get pounds force multiply by 1.356 to get Joules and solve from there.
This is completely incorrect. Force and kinetic energy are completely different and you can not calculate kinetic energy or force based upon one or another simply by multiplying by some magic number. Force and kinetic energy are related in a different way, and converting one to another would require differential and intergral calculus knowledge. From what I know, your conversion coefficient is for converting foot-pounds force to joules, which actually is a measure of work or energy, so you ironically are "converting" from energy to energy.

Even still, you do not multiply by the "area of the barrel" to convert pressure to force. Pressure = force / area, yes, but that is the surface area of the projectile in this case. One side of a sphere has more surface area than a flat circle, which means that a sphere would experience more force than a flat circle.

I'd suggest against using calculations unless you're quite well versed in them. I myself am fairly well versed, but I still never use them due to the vast inaccuracies you'll get under these simplified models. Focus more on building and less on mathematical models at this point.

If you are interested in the mathematical side, perhaps joanna could offer some help to you? I'm under the impression that she is extremely well versed in projectile physics, even fluid air resistance on a projectile.

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Lightbulb41
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Post by Lightbulb41 » Sun Jun 04, 2006 6:17 pm

In regard to converting energy to energy I'm converting from English system to SI so that I can use it in a SI equation so it's safe to do. All this topic was for was to show what I was thinking of building since I had the mathmatics to back it up I wanted feedback on how I should go about this. I didn't want this to become a mathmatics/science debate. Excuse me though. Just wait for a little while... I am going away to work and I'll be able to post once a week so then I can show what is going on. :cool:
You can get more with a kind word and a gun than just a kind word.

I trade mobility for firepower. Accuracy for strengh. And Stealth for Intimidation.

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SSCBen
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Post by SSCBen » Sun Jun 04, 2006 6:58 pm

Umm, now you're switching what you said up? It's perfectly fine to be incorrect. You were converting pressure to force to energy incorrectly. The last bit can be used to convert SI pound-feet to joules, but you did not even have the numbers in the pound-foot unit. I then explained that you'd need to use calculus to convert force to energy so you could use the pound-foot unit. That's what I was saying.

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Lightbulb41
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Post by Lightbulb41 » Sun Jun 04, 2006 7:05 pm

My apoligies... I didn't read correctly Unfortunatly I'm only in 9th grade and I haven't take Calculus at all I'm only taking Trig
You can get more with a kind word and a gun than just a kind word.

I trade mobility for firepower. Accuracy for strengh. And Stealth for Intimidation.

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