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Thanks to whoever made a cleaner version of my free body diagram. Looks great, except that those x and y indices on v and F(sub a) are subscripts, not exponents. I'd really be grateful if you could just touch it up for me. Thanks. -- Random nick 05:16, 5 February 2006 (UTC)
its good too B madan11 ( talk) 16:24, 22 February 2016 (UTC)
It's a high school problem... solved... (No trolling intended... just being clear) -- euyyn 22:54, 25 July 2006 (UTC)
The force due to air resistance is not directly proportional to the velocity of the object, it is directly proportional to the square of the velocity of the object. Imagine the object moving through air at v, the object is colliding with n air molecule per unit of time t. Each collision will exert an average amount of impulse per molecule i. The force of drag on the object will be ni/t=drag. If the speed is increased to 2v, the molecules strike the object twice as hard, imparting approximately 2 time the impulse 2i per impact. At the same time, twice the number of molecules 2n are hitting the object in the same amount of time t. (2i)(2n/s)=four times the drag. Unfortunately, this means that someone who is better than me at WikiMath formatting will have to redo the differential equations. 69.76.242.48 06:56, 13 October 2006 (UTC)
h = launch height
d = flight distance
θ = launch angle (slope)
m = coefficient
You'll get a pretty trajectory (without air resistance in mind, of course) of your favorite projectile on a graphical calculator. Is this awesome, y/n? Well, at least this kind of math is fun. -- nlitement [talk] 01:16, 23 September 2007 (UTC)
I tagged the Trajectory of a projectile with air resistance section with {{ expert}} instead of {{ cleanup}}. The section doesn't require cleanup in terms of the Manual of Style, but it does have a series of equations that to a person unfamiliar with topic (e.g. me) doesn't seem to make much sense and/or offers no explanation as to what is occurring. They seem to be just thrown in together. Barkeep Chat | $ 13:30, 10 June 2008 (UTC)
9.81 ms-2? Why is the exponent -2? Shouldn't that be 9.81 ms2? I'll fix. — Preceding unsigned comment added by Petercorless ( talk • contribs) 20:49, 19 February 2009 (UTC)
I propose to remove from the Trajectory of a projectile with air resistance section the lengthy explanation of why k has the units of N*s/m. It is obvious from Fair = k*V (if Fair has the units N and V has the units m/s then k must be N*s/m). My opinion is that the extra explanation is overkill. Skorkmaz ( talk) 17:46, 29 October 2009 (UTC)
It would seem that in the air resistance section it puts g at -9.8 ms-2, shouldn't it be positive? I checked it out (in Garry's Mod of all things) and it came out with a negative value for an angle when calculating the angle of reach. 219.89.205.8 ( talk) 05:24, 28 November 2009 (UTC)
Firstly, this article only deals with when drag is proportional to speed, which isn't so since for faster speeds F=cv^2. Secondly, the article doesn't deal with how you account for the curvature of the Earth and Coriolis force and centrifugal force. I could add how to deal with F=cv^2, but curvature of the earth depends on where you are, same thing with Coriolis force and centrifugal force. I could add how one accounts for that, but it's going to take some time, maybe during the spring break. So if someone else wants to make additions, please feel free. Pseudoanonymous ( talk) 22:46, 17 January 2010 (UTC)
Maybe the text of the lead section isn't correct. I believe that a ballistic re-entry trajectory for a spacecraft does include drag, but not lift or propulsion. Terminology for artillery projectiles may be different. Martijn Meijering ( talk) 22:34, 15 February 2011 (UTC)
Is the max. height always 1/4 of the range when the angle is 45 degrees and air friction is being ignored? -- SamForestell ( talk) 22:32, 20 November 2011 (UTC)
-- 184.21.215.174 ( talk) 18:23, 14 March 2013 (UTC)
the animation claims a "max angle" at 32 deg in a vacuum. But there is no x acceleration so the projectile will continue to increase indefinitely at the same rate, irrespective of initial angle. — Preceding unsigned comment added by 149.241.5.104 ( talk) 04:09, 26 March 2013 (UTC)
In both the animation and the diagram near the top of the article, it is stated that the launch speed (not x-velocity) is constant for all launch angles. By Pythagoras:
Since Vl is constant, the square-root must be constant, so the larger Vy is (steeper launch angle) the smaller Vx is. When the launch angle is small, Vy is small, so the projectile quickly drops to the ground. It doesn't travel far even though Vx is large. As the angle increases, Vy increases, so the projectile remains in flight for longer and travels further. But when the launch angle is large (more than 45 degrees if the heights of launch and arrival are the same), Vy is large, so the projectile remains in flight for a long time, but Vx is small so it doesn't go very far. In the final case where the launch angle is 90 degrees, Vx is zero, so the projectile goes straight up, then straight down, landing where it started. DOwenWilliams ( talk) 15:08, 26 March 2013 (UTC)
I did the derivation of the 'Angle required to hit coordinate (x,y)'. I belief that there's a missing in equation (2f) and it should look like that:
Can somebody confirm that? DigDra ( talk) 23:59, 1 March 2014 (UTC)
I think the derivations show too many steps, Angle \theta required to hit coordinate (x,y) is quite explicit, same for talking about units for friction.
The Catching section uses a different absurd convention, it use a for 'altitude' and h for velocity. — Preceding unsigned comment added by Timetraveler3.14 ( talk • contribs) 22:50, 15 November 2014 (UTC)
Hello fellow Wikipedians,
I have just attempted to maintain the sources on Trajectory of a projectile. I managed to add archive links to 1 source, out of the total 1 I modified, whiling tagging 0 as dead.
Please take a moment to review my changes to verify that the change is accurate and correct. If it isn't, please modify it accordingly and if necessary tag that source with {{
cbignore}}
to keep Cyberbot from modifying it any further. Alternatively, you can also add {{nobots|deny=InternetArchiveBot}}
to keep me off the page's sources altogether. Let other users know that you have reviewed my edit by leaving a comment on this post.
Below, I have included a list of modifications I've made:
Cheers.—
cyberbot II
Talk to my owner:Online
16:10, 5 July 2015 (UTC)
I removed the second part of Trajectory of a projectile with air resistance as it was completely wrong. In particular, it purported to have a solution for a problem which is known to be unsolved, see " Comments on some recent work by Shouryya Ray" The author makes the incorrect assumption that motion in both axes is independent, which is false for quadratic air resistance since the differential equations of motion are coupled. The first part also needs some work with citations, but it is mathematically accurate. Duxdx ( talk) 01:26, 23 November 2015 (UTC)
If projectile is fired on an incline/slope of beta (angle between -pi/2 and pi/2) then range is (v^2/g) (sec^2 beta sin(2 theta - beta) - tan beta sec beta). Perhaps this could be included in the article (with overview of derivation).
In the derivation for the "Trajectory of a projectile with air resistance", high velocity case (air resistance proportional to the square of velocity), there is a mistake in the components of the force, which also leads to wrong conclusions/calculations after that. In the article the force is taken to be when it should be . Obviously the air resistance should be in opposite direction of the velocity vector which is only the case for the second form of the force. bamse ( talk) 11:38, 24 March 2017 (UTC)
![]() | This redirect does not require a rating on Wikipedia's
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Thanks to whoever made a cleaner version of my free body diagram. Looks great, except that those x and y indices on v and F(sub a) are subscripts, not exponents. I'd really be grateful if you could just touch it up for me. Thanks. -- Random nick 05:16, 5 February 2006 (UTC)
its good too B madan11 ( talk) 16:24, 22 February 2016 (UTC)
It's a high school problem... solved... (No trolling intended... just being clear) -- euyyn 22:54, 25 July 2006 (UTC)
The force due to air resistance is not directly proportional to the velocity of the object, it is directly proportional to the square of the velocity of the object. Imagine the object moving through air at v, the object is colliding with n air molecule per unit of time t. Each collision will exert an average amount of impulse per molecule i. The force of drag on the object will be ni/t=drag. If the speed is increased to 2v, the molecules strike the object twice as hard, imparting approximately 2 time the impulse 2i per impact. At the same time, twice the number of molecules 2n are hitting the object in the same amount of time t. (2i)(2n/s)=four times the drag. Unfortunately, this means that someone who is better than me at WikiMath formatting will have to redo the differential equations. 69.76.242.48 06:56, 13 October 2006 (UTC)
h = launch height
d = flight distance
θ = launch angle (slope)
m = coefficient
You'll get a pretty trajectory (without air resistance in mind, of course) of your favorite projectile on a graphical calculator. Is this awesome, y/n? Well, at least this kind of math is fun. -- nlitement [talk] 01:16, 23 September 2007 (UTC)
I tagged the Trajectory of a projectile with air resistance section with {{ expert}} instead of {{ cleanup}}. The section doesn't require cleanup in terms of the Manual of Style, but it does have a series of equations that to a person unfamiliar with topic (e.g. me) doesn't seem to make much sense and/or offers no explanation as to what is occurring. They seem to be just thrown in together. Barkeep Chat | $ 13:30, 10 June 2008 (UTC)
9.81 ms-2? Why is the exponent -2? Shouldn't that be 9.81 ms2? I'll fix. — Preceding unsigned comment added by Petercorless ( talk • contribs) 20:49, 19 February 2009 (UTC)
I propose to remove from the Trajectory of a projectile with air resistance section the lengthy explanation of why k has the units of N*s/m. It is obvious from Fair = k*V (if Fair has the units N and V has the units m/s then k must be N*s/m). My opinion is that the extra explanation is overkill. Skorkmaz ( talk) 17:46, 29 October 2009 (UTC)
It would seem that in the air resistance section it puts g at -9.8 ms-2, shouldn't it be positive? I checked it out (in Garry's Mod of all things) and it came out with a negative value for an angle when calculating the angle of reach. 219.89.205.8 ( talk) 05:24, 28 November 2009 (UTC)
Firstly, this article only deals with when drag is proportional to speed, which isn't so since for faster speeds F=cv^2. Secondly, the article doesn't deal with how you account for the curvature of the Earth and Coriolis force and centrifugal force. I could add how to deal with F=cv^2, but curvature of the earth depends on where you are, same thing with Coriolis force and centrifugal force. I could add how one accounts for that, but it's going to take some time, maybe during the spring break. So if someone else wants to make additions, please feel free. Pseudoanonymous ( talk) 22:46, 17 January 2010 (UTC)
Maybe the text of the lead section isn't correct. I believe that a ballistic re-entry trajectory for a spacecraft does include drag, but not lift or propulsion. Terminology for artillery projectiles may be different. Martijn Meijering ( talk) 22:34, 15 February 2011 (UTC)
Is the max. height always 1/4 of the range when the angle is 45 degrees and air friction is being ignored? -- SamForestell ( talk) 22:32, 20 November 2011 (UTC)
-- 184.21.215.174 ( talk) 18:23, 14 March 2013 (UTC)
the animation claims a "max angle" at 32 deg in a vacuum. But there is no x acceleration so the projectile will continue to increase indefinitely at the same rate, irrespective of initial angle. — Preceding unsigned comment added by 149.241.5.104 ( talk) 04:09, 26 March 2013 (UTC)
In both the animation and the diagram near the top of the article, it is stated that the launch speed (not x-velocity) is constant for all launch angles. By Pythagoras:
Since Vl is constant, the square-root must be constant, so the larger Vy is (steeper launch angle) the smaller Vx is. When the launch angle is small, Vy is small, so the projectile quickly drops to the ground. It doesn't travel far even though Vx is large. As the angle increases, Vy increases, so the projectile remains in flight for longer and travels further. But when the launch angle is large (more than 45 degrees if the heights of launch and arrival are the same), Vy is large, so the projectile remains in flight for a long time, but Vx is small so it doesn't go very far. In the final case where the launch angle is 90 degrees, Vx is zero, so the projectile goes straight up, then straight down, landing where it started. DOwenWilliams ( talk) 15:08, 26 March 2013 (UTC)
I did the derivation of the 'Angle required to hit coordinate (x,y)'. I belief that there's a missing in equation (2f) and it should look like that:
Can somebody confirm that? DigDra ( talk) 23:59, 1 March 2014 (UTC)
I think the derivations show too many steps, Angle \theta required to hit coordinate (x,y) is quite explicit, same for talking about units for friction.
The Catching section uses a different absurd convention, it use a for 'altitude' and h for velocity. — Preceding unsigned comment added by Timetraveler3.14 ( talk • contribs) 22:50, 15 November 2014 (UTC)
Hello fellow Wikipedians,
I have just attempted to maintain the sources on Trajectory of a projectile. I managed to add archive links to 1 source, out of the total 1 I modified, whiling tagging 0 as dead.
Please take a moment to review my changes to verify that the change is accurate and correct. If it isn't, please modify it accordingly and if necessary tag that source with {{
cbignore}}
to keep Cyberbot from modifying it any further. Alternatively, you can also add {{nobots|deny=InternetArchiveBot}}
to keep me off the page's sources altogether. Let other users know that you have reviewed my edit by leaving a comment on this post.
Below, I have included a list of modifications I've made:
Cheers.—
cyberbot II
Talk to my owner:Online
16:10, 5 July 2015 (UTC)
I removed the second part of Trajectory of a projectile with air resistance as it was completely wrong. In particular, it purported to have a solution for a problem which is known to be unsolved, see " Comments on some recent work by Shouryya Ray" The author makes the incorrect assumption that motion in both axes is independent, which is false for quadratic air resistance since the differential equations of motion are coupled. The first part also needs some work with citations, but it is mathematically accurate. Duxdx ( talk) 01:26, 23 November 2015 (UTC)
If projectile is fired on an incline/slope of beta (angle between -pi/2 and pi/2) then range is (v^2/g) (sec^2 beta sin(2 theta - beta) - tan beta sec beta). Perhaps this could be included in the article (with overview of derivation).
In the derivation for the "Trajectory of a projectile with air resistance", high velocity case (air resistance proportional to the square of velocity), there is a mistake in the components of the force, which also leads to wrong conclusions/calculations after that. In the article the force is taken to be when it should be . Obviously the air resistance should be in opposite direction of the velocity vector which is only the case for the second form of the force. bamse ( talk) 11:38, 24 March 2017 (UTC)