Talk:Terminal velocity

message: //importance as a mechanics relation between constant and variable forces applied, has many applications in applied physics eg parachute, airplane and ballistic weapon design//
I'm not so sure about the equation on this page.  If I enter the following variables:
m=60 kg
g=9.80665 m/s^2
P=1.2 kg/m^3
Cd=1.3
A=1.5 m^2
I get 24 m/s or about 80 km/hr, nowhere near the 195 km/hr listed in the page.  If I multiply straight across, I do get something close to 195 km/hr.
:A (horizontal cross-sectional area) is much less.--Patrick 02:14, 8 Mar 2005 (UTC)
thread_title:  Untitled section 

message: In the article it says that the A in the equation represents the cross sectional area. Would it not actually represent the reference area ("the area of the projection of the object on a plane perpendicular to the direction of motion"), or is it more or less the same thing? CDowns (talk) 23:54, 7 January 2008 (UTC)
thread_title:  Cross-sectional Area or Reference Area 

message: Was wondering if the mention of the terminal velocity of a bullet in the introduction is a good idea? Although the comment is certainly technically correct, I wonder if it might confuse some people as it referces to the free-fall velocity of a bullet, not the firing velocity which is what people might immediatly think of when they see the comment.  Astaroth5 23:04, 14 December 2005 (UTC)
thread_title:  Terminal velocity of a bullet 

message: The equation was wrong and I rectified it. Someone in the recent history edited the article incorrectly, it was initially right. I think there have been some malicious edits .
*Flossie*
any one know this equation:v=kdpower of n
I can think of a more likely explanation.
Hanlon's razor, Wikipedia:Assume good faith.
--68.0.124.33 (talk) 05:57, 10 January 2008 (UTC)
thread_title:  Equation Error 

message: This must be incorrect; that's how long it takes to cross 120mph(176fps) falling toward Earth in a vacuum. (176fps/32=5.5.) The correct answer is infinity...assuming the guy's TVel really IS 120mph. There are too many variables: weight, build, limb position, clothing, attitude, etc. etc. The time estimate should probably be removed, or changed to say how long it takes (longer than 5.5 sec!) to reach a slightly lower value; say, 110mph. I'm too lazy, or I doubt I'd get the math right; you pick. :) --Shyland 13:36, 27 July 2006 (UTC)
replies: text: You're right.  It usually takes about 10 seconds.  In air, the acceleration is much slower than in a vacuum, so it would take 11 seconds to reach 120mph.  --BennyD 22:53, 25 August 2006 (UTC)
replies: text: I'm going to replace the sentence with one giving several values for fractions of the terminal velocity - giving a precise value for when a limiting value is "reached" doesn't really make sense, after all. TeraBlight 21:39, 6 September 2006 (UTC)
thread_title:  Skydiver example: 120mph in 5.5 sec??? 

message: There is a much simpler way to derive the terminal velocity formula, so I included that.  Look at it and see if it's an improvement -- I think it is.  Kier07 00:12, 10 September 2006 (UTC)
thread_title:  Derivation 

message: "Note that the density increases with decreasing altitude, ca. 1% per 50 m (see barometric formula). Therefore, for every 160 m of falling, the "terminal" velocity decreases 1%."
Per the Barometric Formula page the density change is 1%/80m. And, looking at the terminal velocity formula, I think terminal velocity should change as about half of the density change, so 1%/160m seems correct. So I just edited the 50 to 80...someone who knows, double-check me? Thanks, --Shyland 19:48, 31 October 2006 (UTC)
thread_title:  Changed a number 

message: The information regarding Joe Kittinger is debatable.  Follow his link and it will say that his max speed was 614 mph.  On this page it states it is 770 mph and broke the sound barrier.  Since there is no citation on this page I am going to assume that his page is correct and the suggestion that he broke the sound barrier is debatable. Etvander 22:29, 18 January 2007 (UTC)
thread_title:  Joe Kittinger Incorrect 

message: The current definition is:
In fluid dynamics, terminal velocity is the velocity at which the fluid resistance force (drag force) of a falling object equals the weight of the object minus the acting force due to fluid, which halts acceleration and causes speed to remain constant.
Read it carefully. It says that terminal velocity is the velocity at which the drag force equals the weight minus the fluid force. If I am interpreting this correctly (please tell me if I am not), this implies that the fluid force must equal half the weight. I would also like to see more precise language than "halts acceleration".  —Preceding unsigned comment added by Gamesguru2 (talk • contribs) 05:20, 9 March 2008 (UTC)
replies: text: You're right, the definition is wrong. A more correct definition is in the caption of the picture. 76.209.222.33 (talk) 05:38, 11 April 2008 (UTC)
replies: text:  I was wrong here. The definition is correct.  —Preceding unsigned comment added by 76.251.118.99 (talk) 18:18, 11 April 2008 (UTC)
thread_title:  Definition accuracy? 

message: Using the V= equation, it would be impossible to find the terminal velocity in space, because drag is 0, and then you would have to divide by zero, right?  So how do you find the maximum falling velocity in space, then?  —Preceding unsigned comment added by 24.22.200.164 (talk) 23:28, 13 April 2008 (UTC)
replies: text: It appears that we are running the risk of being excessively relative here.   The terminal velocity for an object accelerating into a black hole, for instance, is still the speed of light.   You cannot be accelerated faster than the speed of light.   216.99.198.22 (talk) 02:20, 8 April 2010 (UTC)
thread_title:  Space? 

message: I think that the proof(s) for terminal velocity are badly arranged. I like the fact that the "Derivation of the solution for the velocity v as a function of time t" has the "show" button and I think that the proof above should have one too. Also the sentence "This equation is derived from the drag equation by setting drag equal to mg, the gravitational force on the object." is pointless and should be deleted.--Hevosen (talk) 20:15, 3 January 2009 (UTC)
thread_title: Rearranging the proof for Terminal Velocity

message: When solved without some of the approximations the velocity is not reached asymptoticaly but in a finite time.--Andrestand (talk) 14:04, 4 November 2009 (UTC)
thread_title:  finite time to reach terminal velocity 

message: Some of the formula tags in this article aren't working.Ordinary Person (talk) 13:51, 9 November 2010 (UTC)
replies: text: Could yoy please point out where exactly? Salih (talk) 16:33, 9 November 2010 (UTC)
thread_title: Broken maths tags

message: According to videos of Baumgartner's jump he only reached a maximum velocity of 729 MPH before he began to decelerate, not 834 MPH, as stated in the article.  — Preceding unsigned comment added by 74.95.43.249 (talk) 01:53, 8 March 2017 (UTC)
thread_title:  Incorrect terminal velocity for Baumgartner? 

message: Hello fellow Wikipedians,
I have just modified 2 external links on Terminal velocity. Please take a moment to review my edit. If you have any questions, or need the bot to ignore the links, or the page altogether, please visit this simple FaQ for additional information. I made the following changes:
*Added archive https://web.archive.org/web/20080805143858/http://hypertextbook.com/facts/JianHuang.shtml to http://hypertextbook.com/facts/JianHuang.shtml
*Added archive https://web.archive.org/web/20080331192517/http://www.loadammo.com/Topics/March01.htm to http://www.loadammo.com/Topics/March01.htm
When you have finished reviewing my changes, you may follow the instructions on the template below to fix any issues with the URLs.
Cheers.—InternetArchiveBot (Report bug) 00:59, 22 January 2018 (UTC)
thread_title:  External links modified (January 2018) 

message: The quote by "The biologist J. B. S. Haldane" seems to say something that isn't true, in contradiction of the rest of this entry on terminal velocity. The quote says the mouse won't be hurt because it has a small surface area. Doesn't the formula show that it will still reach a terminal velocity which will kill it? RockyClark (talk) 19:18, 21 July 2021 (UTC)
replies: text: A mouse will reach terminal velocity, but the velocity will be a fraction of that of a man. I added the rest of the quote to make it easier to understand why a mouse falls slowly. It is all about the ratio of mass to cross-sectional area. Those two changes at different rates as the size of an animal changes.Sjö (talk) 04:02, 22 July 2021 (UTC)
thread_title:  Strange quote says something that isn't true