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what itrajectory in a simpler explanation

The opening paragraph is a pretty simple explanation: the path of an object through space. What sort of explanation are you looking for? Samw 23:47, 1 October 2005 (UTC)Reply[reply]

"To neglect the action of the atmosphere, in shaping a trajectory, would (at best) have been considered a futile hypothesis by practical minded investigators, all through the Middle Ages in Europe."

What's this about? Practical-tminded investigation was not the hallmark of the academics of the Middle Ages and early Renaissance. I would have been nice if they'd written mathematical treatises on, for instance, the proper construction of cathedrals (and it would in principle have been possible from the 13th century on), but they didn't; they left that grubby stuff to the trial and error of engineers. (Three guesses who did write the first treatise on mathematical principles of structures. Hint: he's mentioned in the article.) It's true that they wouldn't consider what happens in a vacuum -- because they knew that Nature abnors a vacuum (Aristotle says so); also, the idea of a vacuum is arguably contrary to religion. We could put in that information to mock the poor old Middle Ages, but I think it's more respectful just to zap the passage entirely. -- Dandrake 08:04, 12 December 2005 (UTC)Reply[reply]

You're encouraged to re-write this verbiage and other verbiage in the article. While I added most of the equations in the article, the introductory text leaves something to be desired. Samw 10:09, 12 December 2005 (UTC)Reply[reply]

I received an email flagging a possible problem in the derivation section of "Uphill/downhill in uniform gravity in a vacuum":

Hi Sam,

I was using your trajectory page(nice work) and I am having a problem with part of the derivation. Look at the equation that is fourth from the bottom of the page (this equation is correct). The factoring of sin(Theta)/cost(Theta) appears to be done improperly in going to the next equation.

Could you check this? If this is true, it would affect your comments on this equation that were made earlier in the article around Equation 11.

Mark

My response:

Hmm, you seem to be correct.

The derivation was my work but equation 11 was taken from equation 1-29 at [1]

And certainly the equation for "Rifleman's rule" is an established concept. But you're right, the derivation seems to be incorrect.

Maybe over the holidays I'll have time to figure it out &/or look up the referenced text. In the meanwhile, I'm going to post this errata on Wikipedia and see if others can help. Thanks for contacting me on this and don't hesitate to update the page yourself!

Sam

If anyone can help sort this out, that would be greatly appreciated! Samw 04:52, 25 December 2005 (UTC)Reply[reply]

Hi Sam,

I have been looking at some references and I think I may know what is going on here. The original reference is known to have many typos (see the book reviews at [2]).

I believe the equation in error should have a cotangent instead of tangent in one of its terms.

${\displaystyle R_{s}=R(1-\cot \theta \tan \alpha )\sec \alpha \;}$

I have simulated this situation using an ODE solver and have verified that this equation works and the one in the article does not.

I think I have been able to derive the "Rifleman's Rule" from this expression. It takes a bit of work.

Assume that a rifleman has sighted in his weapon on a flat surface at some range we can call ${\displaystyle R_{0))$. Because the bullet travels along a parabola, the bore of the weapon will have an angle with respect to the line of sight (LOS). Let's call this angle ${\displaystyle d\theta }$. We can compute this angle using the following equation.

${\displaystyle R_{0}=v^{2}\sin 2d\theta /g\approx \ v^{2}2d\theta /g}$ for small ${\displaystyle d\theta }$.

When the rifleman attempts to shoot uphill with his weapon zeroed at ${\displaystyle R_{0))$, the gun will actually shoot further than expected (as we will see). Assume that when shooting uphill (hill angle =${\displaystyle \alpha }$), the riflebore is at an angle of ${\displaystyle \alpha +d\theta }$ with respect to gravity. We can write an expression for the bullet point of impact up the hill as follows.

${\displaystyle R_{s}={\frac {v^{2}\sin(2(d\theta +\alpha ))}{g))(1-\cot(\alpha +d\theta )\tan \alpha )\sec \alpha \;}$

We can approximate ${\displaystyle 1-\cot(\alpha +d\theta )\tan \alpha )\sec \alpha }$ using a Taylor approximation as shown below.

${\displaystyle R_{s}={\frac {v^{2}2sin(\alpha )cos(\alpha )}{g)){\frac {sec^{2}(\alpha )}{tan^{2}(\alpha )))d\theta sec(\alpha )}$

After some simplification, we obtain the following expression.

${\displaystyle R_{s}=R_{0}sec(\alpha )}$

Because the ${\displaystyle sec(\alpha )\geq 1}$, we see that the bullet will hit the hill a bit further up than where the weapon was zeroed. This range extension can be eliminated by firing the weapon as if it were being aimed at shorter horizontal distance ${\displaystyle R_{correction}=R_{0}cos(\alpha )}$. The following equation illustrates this point.

${\displaystyle R_{s}=R_{0}sec(\alpha )cos(\alpha )=R_{0))$

Does this seem reasonable? Note that the "Rifleman's rule" is an approximation but probably holds well for typical situations.

Mark

Wow, quick analysis. Errata for "Modern Exterior Ballistics" is listed at [3] but it doesn't list the cot/tan mixup. Then again, it's not clear that 1-29 came from this book; I only used the webpage I gave and don't actually have the book. Yes, my own derivation clearly points to cot not tan, so I can't object to you replacing equation 11 in the article with the cot version. Furthemore the immediate discussion afterwards on downhill firing and critical angles for uphill firing applies (though the critical angle equations will have to be adjusted.)

As for derivation of the Rifleman's rule, I unfortunately couldn't follow the "expression for the bullet point of impact up the hill as follows" (though I admit I didn't try very hard). May I suggest a step by step explanation? Since my derivation is unfortunately clearly wrong, I'm fine with you replacing my explanation right in the article.

As a sanity check on the whole affair, I must admit I'm uncomfortable on how a tan/cot mixup in equation 11 can still lead to the same Rifleman's rule. I'm half hoping for a second error in my original explanation of Rifleman's rule so this all makes sense again! Samw 21:27, 26 December 2005 (UTC)Reply[reply]

Hi Sam,

I agree with you that a step-by-step development is required. I am new to the Wikipedia, so I am still learning how to do things, like adding a figure (which this derivation desperately needs). Also, I really should look at the original reference.

Thanks for the errata reference. With a little more time, I will try to put something more complete together.

For now, I have been able to solve my immediate problem with the ODE solver. However, an equation-based solution would be more useful to the general public. blacksheep 00:57, 27 December 2005 (UTC)Reply[reply]

For images and Wikipedia, see Wikipedia:How to edit a page#Images. I'd be happy to help in anyway I can. Samw 01:57, 27 December 2005 (UTC)Reply[reply]

Hi Sam,

I have found the key issue. The definition of ${\displaystyle \theta }$ in the referenced web article ([4]) is different than the definition of ${\displaystyle \theta }$ in your work. With the change of variables and a massive amounts of trigonometric simplification, I have been able to show that your work (minus the one error) and the reference work are equivalent.

I prefer your derivation to the reference article (it uses a coordinate rotation that complicates the algebra). However, the derivation in the reference article does allow a simpler path to the "Rifleman's Rule." blacksheep

Hmm, so in order to derive Rifleman's rule, we have the choice of "massive amounts of trigonmetric simplication" to get to my version of equation 11 first or a complex derviation using Taylor's series expansion? Sigh. Now I know why I've never seen a good derivation of Rifleman's rule.  :-) Thanks for the tidbit. I'll adjust the accuracy-disputed flag in the main article to cover equation 11 as well. There's no rush on resolving this; it's sat like this for months! Thanks. Samw 18:42, 28 December 2005 (UTC)Reply[reply]

Hum, looking through this page there seems to be entirely to much mathematics. Do we really need derivations of formula in an encyclopedia? It might be better just to show key results. See Wikipedia:Manual of Style (mathematics)#Proofs and some of this might be bordering on Original research. Its also giving a bit too much away to students who have this set as home work! --Pfafrich 22:37, 2 January 2006 (UTC)Reply[reply]

I'm flattered you think this might be original research! I assume the concern is with Rifleman's rule itself; not the basic trajectory equations. For Rifleman's rule, I have no objections to this material moving to a separate article on Rifleman's rule. Let me know if that's your opinion, and I can do that. There are enough references that clearly this isn't original research. Thanks to the MoS on proofs; I'll try to isolate the derivations more. However, I think the basic equations and derivations are common enough that it should be included. If it's likely to be assigned as homework, all the more reason that this is common enough to be included! Note that there's still trajectory at an elevation and other variations that are probably beyond the scope of Wikipedia.

Now back to the actual issue that triggered this discussion; any thoughts on the actual derivation of Rifleman's rule? Thanks. Samw 04:01, 3 January 2006 (UTC)Reply[reply]

Its still way to complicated. Heres a simpiler derivation

We know solution is a parabola and hence if we write ${\displaystyle p}$ for the position and ${\displaystyle t}$ for the time we have

${\displaystyle p=(At,0,at^{2}+bt+c)}$

the first and second deriv is

${\displaystyle p'=(A,0,2at+b)}$

${\displaystyle p''=(0,0,2a)}$

At ${\displaystyle t=0}$ we have

${\displaystyle p=0,\ p'=(v_{h},0,v_{v}),\ p''=(0,0,-g)}$

hence

${\displaystyle A=v_{h},\ a=-g/2,\ b=v_{v},\ c=0}$.

Giving eqn of parabola as

${\displaystyle p=(v_{h}t,0,-0.5gt^{2}+v_{v}t)\,}$.

To find the (horizontal) range need to find value of t where z-componant of p is zero i.e.

${\displaystyle -0.5gt^{2}+v_{v}t=(-0.5gt+v_{v})t=0\,}$

hence soln are at ${\displaystyle t=0}$ and ${\displaystyle t=2v_{v}/g}$ subs back in eqn for x component of p gives

${\displaystyle R=2v_{v}v_{h}/g\,}$.

By symetry max height will occur when ${\displaystyle t=v_{v}/g}$ giving

${\displaystyle p=(v_{v}v_{h}/g,0,0.5v_{v}^{2}/g)\,}$

To find derivation in terms of angle of inclination ${\displaystyle \theta }$ and speed of projectile ${\displaystyle v}$ we have

${\displaystyle v_{h}=v\cos(\theta ),\ v_{v}=v\sin(\theta )}$

substituting into eqn for range this gives

${\displaystyle R=2v^{2}\cos(\theta )\sin(\theta )/g=v^{2}\sin(2\theta )/g\,.}$

Note that calling this polar coordinates is not strictly correct, we still use cartesian coords but just express initial conditions in terms of speed and angle.

Really thats all which is needed for entire derivation, the rest can easily be dropped.

To find the rifle mans range we wish to find the intersection of ${\displaystyle p}$ with the curve

${\displaystyle q=(v_{h}t,0,v_{h}t\tan(\alpha ))=tv_{h}(1,0,\tan(\alpha ))\,}$

where ${\displaystyle \alpha }$ is the angle of the hill. (I've parametrised the line by t to make things simpler). Equating the z-components and dividing through by t (OK as t=0 is starting soln)

${\displaystyle v_{h}\tan(\alpha )=-0.5gt+v_{v}\,}$

hence

${\displaystyle t={2 \over g}(v_{v}-v_{h}\tan(\alpha ))\,}$

this gives

${\displaystyle p={2v_{h} \over g}(v_{v}-v_{h}\tan(\alpha ))(1,0,\tan(\alpha ))}$

the length of this is

${\displaystyle R_{s}={2v_{h} \over g}(v_{v}-v_{h}\tan(\alpha )){\sqrt {1+\tan(\alpha )^{2))))$

using the trig identity ${\displaystyle 1+\tan(\alpha )^{2}=\sec(\alpha )^{2))$ gives

${\displaystyle R_{s}={2v_{h} \over g}(v_{v}-v_{h}\tan(\alpha ))\sec(\alpha )}$

Divide by ${\displaystyle R={2v_{h}v_{v} \over g))$ gives

${\displaystyle {R_{s} \over R}=(1-{v_{h} \over v_{v))\tan(\alpha ))\sec(\alpha )}$

Finally

${\displaystyle {v_{h} \over v_{v))={v\cos(\theta ) \over v\sin(\theta )}=\cot(\theta )}$

so

${\displaystyle {R_{s} \over R}=(1-\cot(\theta )\tan(\alpha ))\sec(\alpha )}$

So I agree its cot not tan. --Pfafrich 14:06, 3 January 2006 (UTC)Reply[reply]

Sorry I wasn't clear on my question. Yes, User:Ziggle and I came to the same conclusion on the range for uphill/downhill; that equation 11 in article should contain a cot and not a tan. The outstanding issue is how to derive "Rifleman's rule" given the cot. Given all the references on Rifleman's rule, I'm pretty sure the rule itself is valid. Since equation 11 contains a cot, the explanation in the article is no longer valid. Any suggestions on how to derive Rifleman's rule given the revised equation 11?

I see your point now about the overkill in the various derivations. I'll try to find a home for all this in Wikibooks. Samw 22:20, 3 January 2006 (UTC)Reply[reply]

Its a bit late at night for me, so apoligies if this is giberish. Still trying to workout what riflemans rule is all about, (I know some maths but nothing about shooting). One point did occur is that the a good aproximation is ${\displaystyle \sin(\theta )\approx \theta }$ when angle is given in radians. This is a bit more acurate than just saying its zero. This may help some.

I'll have a bash at tiding up the main article tomorrow. Yep it does seem like wikibook would be a good home. --Pfafrich 00:22, 4 January 2006 (UTC)Reply[reply]

Hi folks,

When Sam and I were discussing this page, I did put together a derivation of the rifleman's rule based on this cotangent version. I have not updated the article because I was concerned that it was a bit much. I have put together a web page that summarizes my work([5]). I am sure it could be simplified, but it reflects how my head was working.

blacksheep 22:49, 5 January 2006 (UTC)Reply[reply]

I like it, I certainly some pictures would go well in the main article. A little quible with using ${\displaystyle d\phi }$ d normally stands for infinitesimal derivative. Here you really using it as a difference ${\displaystyle \delta \theta }$ might be more appropriate, as ${\displaystyle \delta }$ is often used for differences.

I've not looked through the maths yet. I'll give it a ponder.

Maybe it time for a new page Riflemans rule, it seems excessive for trajectories. Possible wikibooks? --Pfafrich 00:49, 6 January 2006 (UTC)Reply[reply]

Hi folks,

I have never written a Wikibook, but would like to learn. The "rifleman's rule" does seem to be a well known rule that is missing a good, generally available derivation.

blacksheep 04:35, 6 January 2006 (UTC)Reply[reply]

User:Pfafrich is proposing a Wikipedia article. Just click on the red link Rifleman's rule when you're signed in and start typing! Unless the article is overwhelming in its detail, I don't think it needs to be moved into Wikibooks. Wikibooks is for materials deemed outside the scope of a general encyclopedia. Hopefully tomorrow I'll have time to move the attached multiple derivations for general trajectories over to Wikibooks. I was thinking of adding it to [6] and cross-referencing it here. Samw 04:57, 6 January 2006 (UTC)Reply[reply]

I've added a reference to Wikibooks in the main article. Samw 15:34, 6 January 2006 (UTC)Reply[reply]

Theres a reciently created page Rifleman's rule wondering if we should trim this page a bit, to save repeated material. --Salix alba (talk) 14:32, 31 March 2006 (UTC) (was PfafRich)Reply[reply]

Trim this talk page or the trajectory article? The incline fire section of the article needs fixing. Length of the talk page is harmless. Samw 01:06, 1 April 2006 (UTC)Reply[reply]

Hello folks I would like to add this link to the article: Projectile Lab. It's a JavaScript based trajectory simulator I wrote that asks the user to calculate aspects of the projectile. Any objections? — Edward Z. Yang^(Talk) 21:30, 28 December 2006 (UTC)Reply[reply]

Added .— Edward Z. Yang^(Talk) 21:38, 10 January 2007 (UTC)Reply[reply]

Yow. This page is nearly incomprehensible. Consider doing a rewrite using more English and less math.

The math itself is not consistent. Equation 11 shows cotangent of theta; all the following equations use tangent theta. I assume that cotangent theta is correct, since in the limiting case alpha=theta, the range should be zero. Thus, it looks like everything after equation 11 is wrong.

(looking up at the talk before my post, it looks like you already understand this, but did not correct the page)

I think that a difficulty is that you have never defined whether theta is measured relative to the horizontal, or relative to the surface. I will presume that it is defined relative to the horizontal.

I'm going to delete all the sections that have tan theta instead of cot theta, which is the part between equation 11 and the rifleman's rule. I'm leaving in the rifleman's rule, since I assume it's correct, even though it's not clear how it comes from equation 11.

I find this sentence very difficult to follow: Thus if the shooter attempts to hit the level distance R, s/he will actually hit the slant target. "In other words, pretend that the inclined target is at a horizontal distance equal to the slant range distance multiplied by the cosine of the inclination angle, and aim as if the target were really at that horizontal position."

What in the world are you saying here? The article is about how far a projectile travels. Geoffrey.landis 03:32, 19 March 2007 (UTC)Reply[reply]

Yes the uphill/downhill section has problems, hence the "disputed" tag. In the meanwhile, can you revert back to the orignal form of equation 11? The Rifleman's rule article keys off that. By all means re-write; I haven't had the skill (or time) to correct the error in the derivation. Samw 03:28, 20 March 2007 (UTC)Reply[reply]

As originally written, equation 11 was clearly, unambiguously wrong. Why exactly do we want to revert to a "original form" that is wrong? Not having it at all would be better than having it wrong, I would think. If the following material "keys off of" material that is wrong, rather than replacing the right material with wrong material, perhaps we should delete it-- it is only marginally relevant to the discussion of trajectories anyway. (Another suggestion would be to move all the Rifleman's rule discussion from here to the Rifleman's rule page, and just link that article). Geoffrey.landis 14:04, 29 March 2007 (UTC)Reply[reply]

I think the derivation is wrong. Equation 11 is correct per the discussion above. Hence the disputed tag. Samw 22:56, 29 March 2007 (UTC)Reply[reply]

If equation 11 is correct (as you say) then the following equation (which is unnumbered) must be wrong since it has a tangent in the place where equation 11 has a cotangent. So, since we both agree that it is wrong, why do you want to restore it? Again, I would say that it would be better to have no information than to have wrong information. Geoffrey.landis 02:32, 12 April 2007 (UTC)Reply[reply]

Can we consider adding the following to the "Angle of elevation" section? I don't know how exactly to fit it in, but it is an equation solving for angle with differing initial and final heights. g is acceleration of gravity (e.g. 9.8m/s), v is initial velocity, R is desired range, and Delta h is final height minus initial height. It is self-derived and tested, although I am sure someone more important has derived this before. Unfortunately, it would be almost impossible to find an equation like this on google.Geogriffin 23:33, 29 April 2007 (UTC)Reply[reply]

${\displaystyle \theta =\arctan({-R\pm {\sqrt {R^{2}-4A(A-\Delta h))) \over 2A})}$ where ${\displaystyle A={-gR^{2} \over 2V^{2))}$

Go for it! Just clearly define the terms as I'm not sure what you mean given the terse description above. Samw 03:14, 30 April 2007 (UTC)Reply[reply]

...by anticipating the existence of the vacuum...

Calculating as if the object moves through a vacuum is not the same as anticipating the existance thereof. Rather, what is done is ignoring the friction from the air (drag), since the contribution thereof is often small and the formulae are a lot easier without it. Shinobu 11:04, 12 May 2007 (UTC)Reply[reply]

Be bold and by all means reword that section. Samw 13:23, 12 May 2007 (UTC)Reply[reply]

Thus Rs / R is a positive value meaning the range downhill is always further than along level terrain. This makes perfect sense as it is expected that gravity will assist the projectile, giving it greater range.

Isn't the range assisted by the fact that the ground is simply further away? That is, the lower ground gives the projectile more space to drop, which extends the time it is in the air, allowing it to travel further in the x direction. As far as I know gravity doesn't affect horizontal travel aside from limiting air time.

MagLab (talk) 00:21, 1 August 2008 (UTC)Reply[reply]

Apparently I wrote this 3 years ago. See [7]. Agreed. Feel free to rewrite. Also, feel free to fix the math problems in that section. I haven't been able to figure it out these last few years. Samw (talk) 03:49, 1 August 2008 (UTC)Reply[reply]

I would like to have some information about real rocket trajectories. It would be intersteing to see which flight path real rocket systems use. greets, Andreas —Preceding unsigned comment added by 84.128.60.211 (talk) 20:57, 25 October 2008 (UTC)Reply[reply]

I understand presenting the differential equations that model projectile motion in uniform gravity - but wonder of what help such a contrived derivation could of the equation for the trajectory would be of to the reader. Again the derivation presented assumes a lot of background knowedge - such as "path of the projectile is known to be a parabola"! Well perhaps the editor meant the path is "known" to be a parabola.. given that differntial equations that model it is such.. Anyway.. I propose replacing it with a more direct/simple/elegant algebraic or perhaps even a visual one.

Dilip rajeev (talk) 17:49, 9 March 2009 (UTC)Reply[reply]

I am replacing this unnecessarily contrived and incomplete derivation with a simpler one - the reasoning of which I outline below:

Assume the point of launch as the Origin of the right-hand co-ordinate axes - with the x axis along the ground and y along vertical. Seen from a free-fall frame, which at t=0 is at (x,y)=(0,0) - the object would take a path given by y=xtan(thta). The co-ordnates of this free fall frame is given by y=-1/2gt^2 .. where t=x/(v*Cos(thta)). So translating the co-ordinates back to our original frame, we have ....

Dilip rajeev (talk) 17:49, 9 March 2009 (UTC)Reply[reply]

This article, as well as the projectile motion article, needs a lot of work, IMHO. I have written a good bit of material for high-school and first-year college students (AP Physics) on the subject of projectile motion. Perhaps I can find the time to adapt some of those handouts to make a more accessible and better-organized article...

Math derivations are loads of fun, I've enjoyed doing that sort of thing for some 40 years, but not in an encyclopedia article (nor in most published articles). If the derivation is important, summarize it in the text and put the details in an appendix. Keep the ideas flowing, don't bog down with math, especially when writing for a general, non-specialist audience. Last I heard, an encyclopedia is not a math, nor physics, textbook... (although, full disclosure, in "my" article on Experimental uncertainty analysis I did put in a derivation of the POE expression, just because it is central to the article and is difficult to find).

For a far more detailed (definitely) and accessible (maybe) treatment of this material, you might like to go to this site and scroll down to and click on the /nikenuke directory. Once in there, click the /projectilePDF directory and a whole bunch of PDF papers will be available. These papers will provide a lot of analysis, some of which is beyond high-school level, but several of them should still be useful. These are not the handouts I mentioned above, those are separate; these are intended more for teachers, to provide the mathematical background details as needed for class preparation. Also, there's a couple of Java sims on projectiles in the /nikenuke directory- click on the respective HTML files. Rb88guy (talk) 01:59, 25 September 2009 (UTC)Reply[reply]

Thank you for spelling out clearly what many people have been thinking about this article! A properly-written article reflects knowledge, not only of the subject, but also of how articles should be written. We can take them as conventions; we can take them as rules... but we must heed them! The guidelines for how to create a well-written piece in an academic style (or at the very least, aspire to do that) have evolved over many generations—and they are there for good reasons. If you don't know these time-honoured courtesies, then either learn them or leave the piece to be written by someone else! How do you learn them? You look them up! In much the same way as people are expected, by the authors of certain articles, to consult written pieces on Trajectory or laws of motion. Anyone writing anything for consumption by others must learn the basic language needed to write for that audience. Layers of ill-considered, undisciplined drivel, or rivers of equations that are less than rock-solid simply annoys readers of any level of expertise. For those too lazy to learn more than one simple rule, it should be this: Less is more. Nothing will strangle what little gains you make like extraneous reading load. Eliminate it and you have a fighting chance that your audience will ever want to read anything from you again. Aboctok (talk) 20:14, 29 May 2010 (UTC)Reply[reply]

The article stated: "Physically, this corresponds to a direct shot versus a mortar shot up and over obstacles to the target."

This notion is incorrect. There are two initial angles, theta_0_a and theta_0_b, that have the same range. It makes no difference whether a mortar or a rifle fires the shot. The notion of direct versus indirect fires has no place in kinematics--both are examples of projectile motion, and both are governed by the same rules. "Indirect" and "direct" are military terms, not physics terms.

In reality, maximum range for a given initial velocity occurs when the initial angle of elevation is 45 degrees. In that case, there is only one angle that gives the same range, i.e. 45 degrees. At all other angles of elevation from 0-90 degrees, there are two angles which will give the same range. For those two angles, the absolute value of the difference between said angle and 45 degrees is the same. For example, if all else is equal, and one initial angle is 53 degrees, the absolute value of the difference is 8 degrees. Therefore the corresponding angle is 45-8=37 degrees. So both initial angles 53 degrees and 37 degrees would result in the same range.

Back to the mistaken notion: If you fire your rifle at a 10 degree angle of fire, you could achieve the same range by firing the rifle at an 80 degree angle, but have a plunging trajectory rather than a more direct trajectory, though they would both be parabolas. Of course, this all neglects air resistance, but that isn't covered in this section. Further, when you shoot a rifle, your target is not at the maximum range, which is what the angle of elevation section deals with.

Marktaff (talk) 08:28, 20 October 2010 (UTC)Reply[reply]

Wasn't that the point the writer of that edit was trying to make? Mortars are designed to use the lofted trajectory, whereas rifles could in theory use both, although the aiming would be very difficult without design changes? Martijn Meijering (talk) 08:56, 20 October 2010 (UTC)Reply[reply]

The reason mortars are designed to fire at relatively steep angles has nothing to do with the fact that there are two initial angles that yield the same range in a vacuum. Mortars use such steep angles because they are actually used in an atmosphere, not in a vacuum as these equations assume. In an atmosphere, max range for a small, light projectile (like a bullet) occurs at less than 45 degree elevation, while max range for a large, heavy projectile (like a mortar) occurs at more than 45 degrees. This fact (the effect of the size and mass of the projectile in an atmosphere) is why mortars and rifles typically fire at different elevations. Marktaff (talk) 11:35, 21 October 2010 (UTC)Reply[reply]

That doesn't contradict what I said, does it? It's interesting to know that one type of gun uses one type of trajectory, and the other type of gun uses the other type of trajectory. We must not mislead people as to the reason and the explanation you gave is helpful. Martijn Meijering (talk) 16:27, 21 October 2010 (UTC)Reply[reply]

Agreed. The pedagogical objective is to demonstrate there are 2 ways to hit the target. If mortars & rifles and direct/indirect are confusing to a pure physics student then that's fine to remove. But please do clearly state there are 2 physical trajectories that are possible. i.e. These are not phantom math solutions. Samw (talk) 13:37, 20 October 2010 (UTC)Reply[reply]

It isn't that mortars and rifles are confusing. It is that the mortar analogy just doesn't hold true. Both mortars and rifles can, and in fact are, used to provide plunging fire on targets behind cover. In my opinion, it is bad form to use atmospheric examples (and then, incorrectly) when trying to explain a phenomenon that only occurs outside any atmosphere, so no wind or drag, and constant gravity. I'm not opposed to the two trajectories in a vacuum bit being fleshed out, but we shouldn't use analogies that convey false information. Perhaps that is a spot where a graphic would do wonders? Marktaff (talk) 11:35, 21 October 2010 (UTC)Reply[reply]

The range section ties the range to sin(2theta) is this right? Why 2? Sorry for lack of symbols.

As an impartial observer who just happened upon this page, I wonder if we still need to have a warning that the factual accuracy is disputed for Equation 11. The derivation provided in the Uphill/downhill in uniform gravity in a vacuum section is based on well-known physics principles, and the math is easy enough for an advanced high school student to comprehend. The warning is not meant to last forever, and it has been almost 10 years since the warning was put in place. I vote to remove it. Jdlawlis (talk) 01:34, 15 October 2017 (UTC)Reply[reply]

motion of the arrow of dartboard M shoaib Akhtar (talk) 16:56, 19 October 2019 (UTC)Reply[reply]