Projectiles

Dear steffio,

I hope we do not always ridicule meagre questions but as long as the original question shows intent to learn or a real need to know, we always try our collective best to help.

Your question is not meagre at all as ballistic science is a very demanding field and I know not nearly enough about it to give you a fully correct and scientific answer....however, since you are young (don't know how young) I assume you like building things and playing with it as an experiment.

Make yourself a small launcher and try it out as an experiment to prove it to yourself for real

Take a plastic pipe like electrical conduit. About 20 mm in diameter or so, fix a launch device on one end like an elastic band shaped as a catapult over the open end. Find a projectile that fits loose but snug.

Now you need to figure out a way to have the projectile in the pipe and let it stick out a bit so the hammer of the catapult can hit it always with the same force.

Device an angle readout for the barrel and do some repeat shots as a trial. This will show you the error between each shot.

Now repeat it with angle intervals and take the means of each trial. Put them in a graph against reached distance and you will see a parabolic line with the furthest distance at the optimum angle.

Have fun and you will always be welcome to ask details for the build if you get into trouble. Don't come back here and complain if your mum slapped you around the ears after you broke her antique vase, safety first always.

Good to have you here.

Make sure you pull the elastic back the same distance every shot so the force will be equal.

Also you can repeat the series for a different projectile and a different elastic band. See if all optimum angles are the same!

Thankyou so much for your snappy reply! Much much appreciated.

But just a little more...I'm actually in my final year at school. I have already conducted a projectile launcher and have gotten results that show the general trend that as the launch angle approaches 45 degrees, range increases. However i am supposed to show why i used my hypothesis that is that as the launch angle approaches 45 degrees range will increase. My teacher has asked how i know its not 45 or 46 degrees. i told him "Because it is! It's common sense!" But he told me that's crap. So i've got as far as trying to deduce an equation for it and got x(displacement)=(u(initial velocity)(cos(theta)*sin(theta))/(4.9) ....i don't know if this means anything at all but, do you suppose you could give me a nudge in the right direction?? Sorry for the pestering and thanks again!

If you search for ballistics you should find the info you require.

In actual real terms...45 isn't always the optimum.... If you look at the Masters Golf on TV today you will observe that the longest shots with a driver are much lower than 45 degrees, this is due to aerodynamics and spin of the ball. I believe that an arrow requires an angle slightly less than 45 too.

Theory and practice are not always in agreement.

Having said that... over short distances with a non spinning/nonflying projectile 45 is probably a reasonable bet
Del

Hi Del the cat (cool name),

just wanted to say thanks for your help and time answering a confused kids questions, very much appreciated. Engineers are awesome. Think i might become one now. Ta!

steffio

Theory and practice do not agree in 2 situations:

- the experiment is not done correctly with respect to the assumed theory

- the used theory is not the right one for the done experiment.

I mentioned in a previous comment and mention again : theory is condensed practice.

Now to the problem itself. The angle can be explained in a quite easy manner under assumption that drag is small i.e. velocities are low.

Before giving the explanation i would like to know what is the mathematical level the student has. This will impose the explanation as form.

Hello steffio

If you click on http://www.snipercountry.com/ballistics/index.html

You will find some excellent links, along with downloadable ballistics software, some free.

Cheers and Kind Regards....

45° is the optimal angle in a vacuum and, if you know a little calculus, you can prove that by setting dx/dØ = 0. Once you add air you get a slightly different number due to drag, spin, dimples, etc. Good question and I like the way you responded to answers; come back again.

wow gee thanks for all your help everyone! To TVP45 particularly-you mave have just solved my puzzle. Thanks very very much...

p.s feel free to read my post on "What makes engineers so cool" and add your reasons to the list started. cheers!

Hello Steffio

To test TVP's theory, you really need to go to the moon to try it! Right now is the perfect time for you to start working toward that goal! Good luck and keep up the good work!

Dick

Just an historical physics note:

Galileo used a 45 degree incline to calculate the rate of fall of objects in free-fall. he came very close to 32ft/second per second (32ft/second^2).

The whole Tower of Piza thing is a myth.

Orpheuse

Here's a cool game from the bottom of the newsletter http://www.globalspec.com/Trebuchet/?frmtrk=CR4digest

Or

http://www.globalspec.com/BrainStrainer/?frmtrk=CR4digest

Neither game will answer you question, but are fun with physics

You should be able to prove the obvious analytically. First make two assumptions, there is no air drag and for the distances concerned the earth is flat.

When you fire your cannon the projectile has two components to its velocity, one is vertical and the other is horizontal. I would think that the horizontal component will be zero when the projectile reaches the ground.

Using simple trigonometry you should be able to calculate the horizontal component and the vertical component. As the projectile rises the force of gravity will cause the vertical component to become zero, it will then start accelerating the projectile downwards until it strikes the ground. Given the initial velocity of the projectile you should be able to calculate the time required for the projectile to reach its maximum height and return to the ground. Using this time and the horizontal component of the velocity you should be able calculate the distance the projectile travelled.

Now you just have to show which value for theta results in the maximum distance given the assumptions that you have made for your theoretical projectile. This is a trade off, you can maximize one component of travel, either vertical or horizontal at the expense of the other.

Get out your spreadsheet and try graphing range vs theta and see where the maximum is. Generate an expression that calculates the range as a function of theta assuming that the muzzle velocity is a constant and then use normal mathematical techniques to find the value of theta that will give the maximum for the expression.

Good luck on your assignment

Good answer, But rather than (erroneously) assume the Earth is flat, suggest you start with a straight line drawn between target and launch device. Always better for the artillery to be on the high ground :)

Dear The Curious One,

Thankyou so much for your detailed answer, i understand how to get the range as function of air time by horizontal velocity and have since tried to get an expression for range as a function of theta but that is the precise bit i get stuck on... so far i've only got range= (u(cos theta * sin theta))/4.9 which i don't know what to do with. (probably nothing). Can anyone work out an equation to get range as a function of theta??

Deadline=Thursday...aghhh!

Thanks again uber cool engineers!

So, sounds like you need that great trig identity (at last you get to use it for something!) of

2sinΘcosΘ = sin(2Θ)

Oh, and you don't really need to worry about the constants like 1/4.9 etc since you're looking only at maximum range, not actual range. So, if you're OK with what I said, you could say that range = sinΘcosΘ and that should make it obvious.

Range is given as u cos (theta) *time. The time it takes the projectile to fall from its maximum height to the horizantal ground level is the same time it takes the projectile to travel its range.

i.e. u t cos theta = u t - 1/2 g t^2 ----(1)

the other equation needed is

(u sin theta )^2 = (u cos theta )^2 -2 a s ----(2)

where:

u projectile velocity at angle theta and a is the horizantal acceleration which is zero (0) at the turning point of the motion.

From (2) therefore you have

u^2( sin^2 theta -cos^2 theta)=0

u^2 cannot be zero therefore

sin^2 theta -cos^2 theta = 0

or

-cos 2 theta =0

the cosine function is an even function, therefore

-cos 2 theta =cos 2 theta = 0

or 2 theta =90 degrees

therefore theta =45 degrees.

Thank you Neville Tomlinson for the maths exercise.

Simple logic would establish (without using any maths at all), that 45 degrees inclination upwards from the horizontal is the optimum angle for the greatest range, if the complete range is across a level and flat surface..

Kind Regards....

Hi Steffio

Maybe you already know that the projectiles motion can be analyzed in two movements independent from each other: One horizontal and one vertical. This is analysis is typical for most kinematics problems. In our case, the horizontal movement, is at a constant speed (assuming there is no air resistance) - as there is no horizontal force applied to the projectile - and is described by x = v_hort (1). The vertical motion is on the other hand governed by the acceleration caused by earth's gravity (=g), therefore it is:y = v_vert - 1/2 gt² (2)

The projectile will reach the ground after the end of the trip when y=0. I bet you can find the time it takes for this, by using (2).

During this time, the projectile traveled distance x=v_hort and by substituting the above found time, you get the point on earth that the projectile will land. This is what we want to maximize.

From now on, you just need trigonometry. First of all, you need to know that v_ver=v.sinθ and v_hor=v.cosθ . I'm sure you know it already. Then, you will need this trigonometric identity: sin(2θ)=2.sinθ.cosθ

If you do the simple math, you will notice that in the final result, you will have something constant multiplied by sin(2θ). In order to maximize this, we need to find the θ which maximizes sin(2θ). I feel you can find yourself what is the maximum value of sin(2θ) and for which θ we get it.

Use same mass and launch force. Calculate every degree from 0-90 for distance. You can graph the result also. Should only take you 15 minutes.

To answer this simply: To get maximum distance from a projectile, set your angle at 45 degrees; anything above 45 degrees and you run into the [laws of diminishing returns], that is to say that at 50 degrees, your projectile will start falling closer to you. Anything less than 45 degrees and the projectile will not travel as far (given that there is no change in topography or atmospherics).

That may be a bit over-simplified but it does answer the question.

For the equations see tkot and rainbow, they nailed the question to the wall.

Orpheuse

I think that the person who gave you the assignment will be impressed with the analytical approach using the equations and hints that you have been given here.

It is unusual to see so much assistance given to a person with a homework assignment, but I think that all of the posts have been teaching posts rather than doing the actual assignment for you which is often expected.

Good luck and may your teacher, instructor, proffessor or whatever be impressed by your work and may you have learned from the experience, that is the most important part of the exercise.

Dear ALL,

With your combined physics brains and my reading skills i think i've got everything i need!! Thankyou all so very much! I'm just about ready to hand this report in an A+ it, thanks to you guys. THAT is why engineers are so dam cool.

Steffio.

A quick note before you go, I was involved a long time ago as a technician maintaining the many Friden Calculators used by US artillery, the type of people who "care" the most about this subject. Try Googling artillery ballistics for some definitive insight.

Hi steffio,

A meager suggestion, if that's okay. There's no such thing as an "ideal" launch angle in the sense you asked. An ideal trajectory could be anything you wanted it to be...even "straight" up or "straight" forward...depending on target, terrain, intention...objective and subjective things like that. What you seem to want to know is: ...how to calculate the optimum launch angle for maximum range on level terrain and with zero precip and zero wind, and ignoring direction/earthspin. See? ...always important to precisely define your problem.

I launch my rockets slightly into the wind, and typically want them to fly high and come down by parachute.

The best place to launch projectiles is important since in space 45 degrees is the same as 90.

When launching rockets it is good to get aid of the spin of the Earth and why putting your launch pad at the Equator helps.

In my ideal world my rocket returns to me without me having to run around looking for it.