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# Deriving the maximum range and angle of a projectile Answered

In a thread in the Green group, AnarchistAsian and I were discussing his coil guns. I posed the question of what range he could get, and he asked me to go through the physics derivation. Here it is, simplified to require just algebra and trig.

The kinetic energy of the projectile E = ½mv2, gives v = sqrt(2E/m) as the speed at launch. Let θ be the angle at which you launch (θ=0° is horizontal, θ=90° is straight up). Then you can decompose the speed into two components:

vh = v cos(θ) is the horizontal speed,
vv = v sin(θ) is the vertical speed.

Gravity only pulls vertically, so the projectile's vertical speed will be slowed down until it reaches its maximum altitude, then it will fall back until it hits the ground. The horizontal speed will remain constant until it hits the ground and stops. To figure out the range, you need to know the time t that the projectile flies before it hits the ground; then the range is just

R = vht

Energy conservation guarantees that it's downward speed at the end is equal to its original upward speed, just with a change of sign. That also means that the total flight time of the projectile will be half going up and half going down. Once you determine how long it takes to reach the top of its flight, you're done; just double that answer :-)

In the vertical direction, the maximum height

H = vvt - ½gt2

(you need calculus to derive this result). From energy conservation, the initial kinetic energy in the vertical direction, Ev = ½mvv2 must equal the potential energy at the top of the flight, Pv = mgH:

mgH = ½mvv2
H = ½vv2/g

Substitute this on the left hand side of the trajectory expression,

½vv2/g = vvt - ½gt2
vv2 = 2gvvt - g2t2
vv2 - 2gvvt + g2t2 = 0
(vv - gt)2 = 0
vv = gt

So, t = vv/g = v sin(θ)/g is the time to reach maximum height. Double that as discussed above, and you get the range R = 2 sin(θ) cos(θ) v/g.

Work out the angle that gets you maximum range by just plugging in different angle values and finding the one that is biggest.

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## Discussions

Ignore the italic "angstrom" symbol :-( I'bles WikiFormatting processor has some stupid problems with the extended ASCII characters (character codes 128 to 255). It sticks in extra symbols, and then if you try to edit the text to get rid of them, it actually adds even more of them!

Those two statements are supposed to be "Y = 0 degrees is horizontal, Y = 90 degrees is straight up", where I'm using "Y" as the symbol for angle instead of Greek "theta".

> Those are standard Unicode values?
. I dunno. Found 'em at Entities for Symbols and Greek Letters. That page says "Glyphs of the characters are available at the Unicode Consortium," so I'm guessing they are.
.
> ... there are 65,536 possible!
. What do you think about having a spreadsheet for download? Using Fill and C&P, it shouldn't take too long to build - that's what I did for the ASCII chart screen captures.
. I'm guessing that a LOT of the characters are not needed (eg, any pictograph language chars). For use on Ibles, just the English (Roman?), Greek, and Math chars should be enough. Probably ought to include any German, French, &c chars for those times one wants to spell Gotterdammerung or tete-a-tete properly.

German, French and Spanish are already part of the ISO Latin-1 encoding (the table you have :-). They all have proper entity names, as well as numeric values: &eacute; = é &ouml; = ö and so on.

It's the non-Latin characters for which you have to go to Unicode (Greek, Cyrillic, the Cyrillic-derived Eastern Europeans, etc.).

. Does that mean that all I need is a supplement to the ASCII/Latin-1 chart, with the Greek Unicode characters? Or do you think there is more that should be added? . I don't want to be Anglo-centric, but the charts may be too big if the "minor" (eg, Cyrillic) languages are included. With a SS, it wouldn't be a problem. Arrrrgggghhhh! We need tables in the markup! Then the user could copy the entity directly from the page. . heehee Yeah, Latin. Not sure where I got Roman from.

Rock on! Congratulations on this milestone :-) You've got it troubleshot (troubleshooted? whatever...) and ready to go? I look forward to seeing your results.

If you really can get an altitude of several kilometers (I highly recommend launching out to sea, if you can :-), that will be quite an achievement.

well, the parts are on the breadboard, now we just need to make the connections...
The ible is here, it's not very detailed, but the diagram is pretty good.

altitude of several kilometers... Yeah, that makes sense i guess, since a .22 can go pretty far, even if in-accurate, but the thing is, the efficiency of the coil gun...

My dad has some 10awg wire at his lab, enough for me to have several stage's worth of wire... how many stages do you think it would take to get 75% efficiency or higher?

Ack. Now you're getting into electrical engineering, and that's way outside my experience. It seems to me that you have the knowledge to at least search for that sort of information, either here on I'bles, or to recognize "truthy" :-) Google results. Binary guy, westfl, GH, NM, any of those folks probably know more about practical electrical stuff than I do.

Yes, that's true. I can do the basic calculations, and make simplifying assumptions so I can "figure out" how something works, but I don't know electrical stuff at the engineering level. My HEP work is on the detector side, and on data analysis. If I were directly involved with the accelerators, I'd probably be more familiar with projects like yours.

Well, you can do the math yourself -- you are a rocket scientist, after all ;->

Uranium has a density of 19.1 g/cm3 (you can probably neglect the steel coating). Your BB has a diameter of 0.3 cm. Now you have the mass and velocity, so you can calculate the kinetic energy.

Use Wikipedia to look up the physical properties of the material you're shooting at (glass, wood, steel, people). That should give you enough information to convert energy to penetration depth.

Please do. The only place where calculus is really needed is to derive the projectile height as a function of time.

If you're willing to accept that equation "on faith" :-) then everything else is just algebra, plus substituting in the numerical values for your coil gun.

I would definitely encourage you to go through this with your Dad. Try it on your own first, making sure you understand each step; feel free to ask any questions you want here. If you think you've found a mistake, definitely post it here -- just because I wrote it down doesn't make it true :-)

Heh. Okay. I guess the two equations for vh and vv you can "take on faith" for now. Your calculator will have buttons labelled "sin" and "cos", which you can use to plug in different angles (Y values) at the end of my writeup. You can explore the range as a function of those angles and see where it is maximized.

When you take trig (or look it up on Wikipedia) you will see why those two equations work the way they do.

calculators? no, we always do by hand, when possible...

Anyways, my dad just finished showing me how to do it, (after waiting for him, and then looking all over the place for those dry-erase markers) on a dry erase board we got a long time ago...

Well, here's what i got: and they areWAYoff, since the air resistance isn't acounted for...

Distance up and down is 50,000 meters, =-o so then range is 100 km?
And... as for the angle, 45 is always the way to go, right? i thought so, and my dad said so...

Wow, that is a huge amount of kinetic energy you're going to put out. With a vertical height (at 45°?) of 50 km, your half flight-time is 101 seconds, at an initial vertical speed of 990 m/s -- is that what you got? Your total time downrange is 202 s, and at 45° your horizontal and vertial velocities are equal, which means a range of nearly 200 km.

With those numbers, all of my approximations are very approximate :-) You not only have to account for air resistance, you have to account for the different air resistance at 50 km altitude (that's halfway to low orbit!), and you probably need to take account of the Earth's curvature to compute the true downrange distance.

The 45° angle is the optimal solution for maximum range. I was hoping that you'd actually go through the math -- use the range formula at the end, plug in different angles to compute sin(Y)cos(Y), maybe even make yourself a graph of range vs. angle. Then you could see the 45° solution instead of just being told :-)

This was a really good exercise; thank you! I hope that you got something out of it.

oh, we got flight time of 100 seconds, coz we rounded.less work, fairly accurate right?
990 m/s,uhh, well that's about twice what it was. going to it's highest point, it goes 25,000 M.
I calculated about 569m/s. HOPEFULLY *crosses fingers*, we can get %70 of that speed...still a lot, but i have a "need for speed"

50km is way off though, the air resistance would lower that more than 10 fold... =-(

Well, we already concluded 45° was the best angle, but he still showed me how to calculate it. So it was 770m/s in that direction... But that was with 1000m/s, that i thought was the speed, but it was a mis-calculation =-(

ok, so anyways, the capacitor bank has 4800uf, at 450 volts.

Rounding off is perfectly correct, AA. I do it all the time, and it is an excellent tool for you to learn. We call it an "order of magnitude" or "back of the envelope" estimate. If I were doing this in my head, or pencil and paper, I'd use 10 m/s2 for gravity instead of 9.8, and so on.

About the 990 m/s vertical I asked about, maybe I got it wrong? It sounds like one of us had an extra factor of two floating around :-)

In any event, that is one heck of a gun you want to build. 972 joules of stored energy (U = ½CV2); call it 1 kJ. If you're estimating 1 km/s launch speed, I guess your projectile is only going to weigh about 2 grams? That's not very big; steel has a density of about 8 g/cm3. You're going to launch something like a ball bearing?

hmmm... 990, maybe you used the 50,000 as to the top, instead of up and back, i think that's the most likely...

close, not 2 grams, 3 grams. It's a steel rod, 3/16 inch, or 1/4 inch, i can't decide... 1/4 will have more stopping power, i'd say... both at 1inch long.

972 joules? oh crap, i got 486... *re-calculates* i still got 486! crap, somethings wrong... ok ok, so i'm going to tell you everything i did:

i squared 450 and got 20250, then multiplied by 0.002400, and got 486.
Then i divided that by 0.0015, and got 324000, then square root was 569 m/s.

hmmmm, your answer is always twice what mine is, did i make any mistakes?

That's where you need calculus (among other things). Air resistance is a frictional force, so it acts directly opposite the direction of motion. It's roughly proportional to velocity. What you have to do is write down a function for your velocity vs. your position along the trajectory (path length), then you add in the air resistance at that velocity acting to slow you down. Then you integrate (calculus) the whole function to work out the resulting actual trajectory. It gets pretty complex, and the text interface here couldn't handle the math notation anyway.