# Calculating Drag When Firing Potato Launcher and Design Specifications

I'm interested in making a pneumatic potato launcher, I've found plenty information on that but I want to hit a target 255 meters away at an angle of 45 degrees. I calculated that I would need an initial velocity of approximately 50 m/s without wind resistance. The object being launched is an average sized potato under normal weather conditions. With wind resistance, what initial velocity will be required to hit the target? Please explain the process by which you came to the answer so that I may use the same method in future endeavors with different angles and distances. Also, just how much pressure would be required to launch a potato and achieve that speed, and to what specifications would the launcher need to be built i.e. how large the gas chamber will need to be, length of barrel, etc...

## Discussions

8 years ago

You can do this whole problem yourself, if you make some approximations, and are willing to do some research (rather than waiting for someone else to do your work for you). It is

notsimple, but it is tractable. If this is a classroom project and you expect us to do your work for you, then you're hopefully out of luck.First, the air resistance issue. Potatoes are irregularly shaped, and each one is unique, so trying to get an exact value is silly.

If you have access to a small wind tunnel (I don't find an I'ble to make one, so check Google or Wikipedia), then you can measure the drag yourself with a sample of different potatoes, and take an average (or largest value).

Alternatively, you could do the calculation by assuming the potato is a smooth sphere, and looking up the answer. Either way, keep in mind that drag is proportional to velocity, so if you just take the drag result and increase your launch velocity, the resulting drag will be even higher. The process does converge (obviously :-), but might need to iterate your calculation a couple of times.

Second, pressure to get launch velocity. This is trivial arithmetic based on definitions and unit conversions. What is the cross sectional area (A) of your potato and launcher (they have to be the same, obviously)? Let P be the unknown pressure; then the total force on the potato is just F = P*A (with consistent units).

Given the mass (m) of your potato, the initial acceleration (a) is found from F = m*a. The length of the launcher (l) determines how long the acceleration is applied (l = 1/2 a*t

^{2}), and then v = a*t gives you the muzzle velocity.Algebra (or calculus, if you want to take into account the pressure drop as the potato moves down the barrel) is your problem, not mine.

Reply 8 years ago

Build your own wind tunnel.

Without having an immediate clue how to calculate it, isn't the pressure issue more complex than you make out? As the potato moves up the barrel, the volume "behind" it will increase, so the pressure drops as the potato moves.

And what about friction between potato and barrel? That will vary wildly with fit and moisture content...

I suppose what I'm saying is that this is one of those problems I would prefer to solve by trial and error.

Reply 8 years ago

Those are all complications, and depending on how accurate the author wants his results, he would need to incorporate more and more of them. But you can still get very useful "back of the envelope" results by ignoring them (this is the physicists' spherical-cow argument).

8 years ago

It's not a classroom project at all; it's a completely personal thing I'm doing for recreation. I recently learned about projectile motion in physics class but, unfortunately, we haven't covered wind resistance yet. I'm in the process of constructing my own potato launcher and I want to be able to fairly accurately hit targets at predetermined distances. So I basically just wanted to know how I would factor wind resistance into my calculation. Thanks for the information you provided though, it was pretty helpful!

Reply 8 years ago

:-) Great! You may not have seen the absurdly huge number of questions posted here which essentially boil down to "please do my homework/exam/project for me." Sorry for making an unwarranted assumption. This is a really good example of a very practical problem for which you can compute a reasonable (but approximate!) solution. To factor in wind resistance, the basic idea is that it's a frictional force. It's going to be proportional to the area facing into the wind, and increases with increasing velocity. Wikipedia ought to have the basic equations.

8 years ago

...Kelsey!Reply 8 years ago

Hee, hee! You posted this while I was writing my analysis. I'm not about to give away the answer, but this did seem like an opportunity to give some hints on

howto solve it.It introduces at least three broad physics concepts, along with the general problem-solving technique of simplifying approximations. That should be more than enough for Mike714 to go the rest of the way on his own.