Talk:Terminal velocity

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Untitled section[edit]

//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)

Cross-sectional Area or Reference Area[edit]

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)

Needs a more complete treatment[edit]

I'm no physicist so I can't do this myself, but this article should answer a few additional questions. For instance, assuming an optimal shape like a cannonball, what is the fastest possible freefall rate? How far must the object fall to attain that rate? As measured at 10 feet above ground, a cannonball released from an airplane at 40,000 feet is falling no faster than a cannonball dropped from an 80th floor window? A person falling from the 50th floor splatters no less than one falling from the 100th? Would terminal velocity be greater on a planet double the size of earth with an identical atmosphere? What if that planet's sun was quadruple our sun's size? Was it Galileo who first figured terminal velocity or someone later? Mass is one of the factors in the terminal velocity equation given in the article-- does this mean mass influences the km/h terminal velocity figure for a given object? If yes, how is this consistent with Galileo's law of falling bodies (that heavier bodies do not fall faster than light ones)? If no, why is mass a factor in the equation? JDG 03:48, 18 Apr 2005 (UTC)

Holy cow. You sure are full of questions. I'll address them as best I can here, and it can be decided later whether they belong in the article or not.

  • First of all, talking about "fastest possible freefall rate" is pretty meaningless. As mentioned in the article, terminal velocity involves a number of factors, including mass and shape. This is because terminal velocity is reached when gravitational force down is equal to aerodynamic drag force up. The heavier an object is, the more gravitational force is exerted on it, which means a greater drag must be applied to balance gravity, which means that a higher velocity must be reached (because drag is related to velocity). So each object will have its own terminal velocity based on its physical properties. Therefore, your questions relating to "maximum possible freefall rate" and "how far must an object fall to achieve that rate" can't really be answered in a general fashion.
  • As for your questions about where an object is dropped from: assuming an object has accelerated to terminal velocity, it will be the same terminal velocity (at a given altitude; drag is dependent on density and density changes with altitude, so terminal velocity changes with altitude) no matter where it was dropped from.
  • If a planet had double the mass of Earth with an identical atmosphere (as unlikely as that seems), yes, terminal velocity would be greater, because gravity would be stronger. As far as the size of the sun...that's not really relevant.
  • As stated, yes, the mass of an object changes its terminal velocity. This is not in contradiction with Galileo's law of falling bodies, because Galileo's law assumes a perfect vacuum where there is no drag.


Missing explanation of Terminal Velocity in situations outside free fall[edit]

This page only explains terminal velocity of an object accelerating in free fall due to the force of gravity. Terminal Velocity is the limit velocity approached due to any accelerating force, not just gravity. If you were to replace g in your equation with the acceleration due to any force it would be more accurate. As much as your example of Newtonian Physics represents the most common occurrence where one would seek to determine Terminal Velocity your posting could be improved by applying the physics of the general case.- SB

Terminal velocity is a phenomenon associated with freefall in an atmosphere. This is how it is used in the physics and aerodynamics fields, so why should Wikipedia be any different? --boeman

The definition doesn't make sense to me[edit]

First of all, can you say that air resistance "pushes an object upward"? Isn't it more correct to say that the air resistance creates friction that slows the object down? Also, what does "equal and opposite" mean? It makes more sense to me to say that the resistance is equal to the acceleration, thereby effectively stopping acceleration. The way the definition reads now, it implies to me that the terminal velocity actually suspends the object in the air. I don't have any idea what is meant by "opposite".

In terms of physics-speak, the article is correct. Wikipedia is an encyclopedia, not a textbook. If you want to learn more, then do so.--boeman

I think the article also requires an example, such as a skydiver or a falling ball.

There are two meanings[edit]

Before the Charlie Sheen movie came out, I don't most people weren't that familiar with skydiving terms, and (where I hung out), "terminal velocity" seemed to mean, literally, the velocity at the end of an object's path.

So, if you fired a probe into an asteroid, or accelerated a particle against a target, the speed at which it actually hit was, literally, its terminal velocity. For a body free-falling in an atmosphere, that final, terminal velocity is obviously going to be affected by atmospheric dragging effects as well as gravity, and the "skydiving" usage might usefully be referred to as the maximum achievable terminal velocity for a particular object, due to the stronger amounts of drag at progressively higher speeds.

"" currently lists both meanings, with the "general" meaning first, (source: Websters abridged).

I think this more general meaning (a final velocity that may or may not be constant) ought to be given at least a small mention on the page.

Terminal velocity of a bullet[edit]

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)

Hi,i've been set a question about what is terminal velocity in my year nine science class at school (i'm 13) and think it is good that it is simply and clearly explained in the first paragraph. However, i would like to know more about the history of how terminal velocity was discovered on the page.What's the speed of Termnial Velocity 120Insert non-formatted text hereÍ

lets get this back in or better have a link to a relevant article or section of an article under ballistics. airplanes fire bullets which reach ground at terminal velocity i guess, also of a bomb. (talk) 22:37, 3 April 2008 (UTC) i have put this info back in, after reading the reference i have been able to clarify , i hope what it is about, it does refer to the terminal velocity ie the doenward speed, of the bullet when falling to earth under gravity. (talk) 00:15, 4 April 2008 (UTC)

I think it is important to clarify that the bullet is fired vertically. This is mentioned in the article referenced but omitted from the citation. While the downward (toward the earth) acceleration from gravity of a bullet is limited by terminal velocity, its speed when fired in a ballistic trajectory could exceed that of the terminal velocity of a bullet dropped from a height equal to the apogee of the trajectory.Jdcook72 (talk) 21:07, 22 November 2011 (UTC)

Equation Error[edit]

I had it pointed out to me that the Terminal Velocity equation is actually:

Vt = √( 2 *m *g / (Cd * rho *A ))

The equation on the article is missing the "2".

Equation Error[edit]

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. -- (talk) 05:57, 10 January 2008 (UTC)

Skydiver example: 120mph in 5.5 sec???[edit]

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)

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)
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)

There needs to be a listing of what units are to be used with each of the variables.

Brick example[edit]

I'm having an argument with my neighbor about moving the pile of bricks beside my house so they don't pose a missle hazard during a hurricane. I wanted to tell him the wind speed required to pick up one of my bricks and fling it through his window so I thought terminal velocity would be the right estimate

m=2.64 kg g=9.80665 m/s^2 P=1.2 kg/m^3 Cd=2.1 A=32 in^2 = 0.021 m^2

I get Vt = 113km/hr = 70 mi/hr. That just doesn't seem right. It seems like a brick should fall way faster than a human body. Xcross 19:28, 30 August 2006 (UTC)xcross

Well, if you mentally simplify the expression a little, there's two variable contributions that need to be considered.
Firstly, there's a fsctor of m/A, i.e. the mass of the falling object per unit area. For you brick, that's about
2.5 kg / .02 m^2 = 125 kg/m^2
while for a human, let's say
70 kg / (1.75 m * 0.4 m) = 100 kg/m^2
In other words, the value is slightly higher for the brick, which means the terminal velocity is slightly increased.
The second factor is a 1/C, i.e. the inverse of the drag coefficient. Interestingly, the drag coefficient for a brick is apparently twice as large as that for a human. I assume it's because of the brick having sharp edges, while humans are a lot more rounded and hence more aerodynamic.
Combining these rough estimates, your value sounds about right. Somehow, our intution when it comes to falling objects is often not very good. TeraBlight 21:35, 6 September 2006 (UTC)

Heh heh, I suppose falling bodies is not the only place where human intuition and physics depart. So back to the argument with my neighbor, it sounds like greater than 70mph wind might suffice to move the brick through the air. But what about picking it up? I suppose you would start with the frictional force on a brick on top of the stack, no?

Well, as soon as the angle of attack of the airflow is even slighly from below the object rather than sideways or from above, friction ceases to play a role. The terminal velocity gives you the windspeed it takes to hold the brick in the air if the wind comes from exactly below (quite a hypothetical scenario, but gives one an order of magnitude idea). At any windspeed less than that, the brick may still be dislodged but will fall to the ground quickly irrespective of the direction of the flow. At higher windspeeds, the brick can maintain altitude under the right conditions. Does that help at all? TeraBlight 07:29, 12 September 2006 (UTC)

I think if the wind is strong enough to pick up even a lone brick off the ground, flying bricks will be the least of his worries. Flying houses will be higher on his list. --Shyland 08:35, 4 November 2006 (UTC)


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)

Changed a number[edit]

"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)


When applying this equation to objects falling through liquids do you have to take account of the bouyancy force when determining the mass or does the density term in the equation already account for this?

Joe Kittinger Incorrect[edit]

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)

Possible contradiction[edit]

In the beginning of the article it's stated that "Terminal velocity varies directly with the surface area of the object confronting with the air or fluid but should not ever be related to the mass of the object", but in the equation the mass is indeed used and it's clear in the article that an object with greater mass has a greater terminal velocity. Is there an error or am I missing something? Hybor 09:04, 6 February 2007 (UTC)

Yes that's an error. Dropping a 1-gal. bag of feathers falls a lot slower then a 1-gal bag of water, even though they will be the same shape and size. They would fall at the same rate on the moon; it's only because the bag of feathers reaches its terminal velocity right away but the bag of water keeps on accelerating for some time. -Jesse —The preceding unsigned comment was added by (talk) 20:16, 8 February 2007 (UTC).
I fixed it. -Jesse

Definition accuracy?[edit]

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 (talkcontribs) 05:20, 9 March 2008 (UTC)

You're right, the definition is wrong. A more correct definition is in the caption of the picture. (talk) 05:38, 11 April 2008 (UTC)
I was wrong here. The definition is correct. —Preceding unsigned comment added by (talk) 18:18, 11 April 2008 (UTC)


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 (talk) 23:28, 13 April 2008 (UTC)

If you divide by a limit close to zero, you'll get almost infnity - in space terminal velocity would be infinite, which is correct, since with no drag, you can accelerate at the same rate for eternity. (And no, even if you accelerate at the same rate forever, you won't exceed the speed of light barrier, BTW) —Preceding unsigned comment added by (talk) 07:51, 9 May 2008 (UTC)

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. (talk) 02:20, 8 April 2010 (UTC)

This article defines terminal velocity of a body to be the maximum velocity experienced by that body in a fluid. It is explained that the fluid drag on the body is what limits the velocity of the body to some maximum value. The concept of terminal velocity in space (ie an environment devoid of fluid) is not relevant to this article. I agree that it is valid to question the maximum velocity of a body in interstellar space, but Terminal velocity is not the place for it. Dolphin51 (talk) 11:37, 8 April 2010 (UTC)

Invincible Cats[edit]

I've heard it said that any animal whose terminal velocity is below the velocity needed to kill it cannot die in a fall. This was in the context of cats in New York, where many cats who fall from 30+ stories survive. I assume they have high drag because they are fluffy. This might make an interesting anecdote for the article if it helps explain a concept. What other animals might be invincible? (Besides flying squirrels.)

Rearranging the proof for Terminal Velocity[edit]

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)

finite time to reach terminal velocity[edit]

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)


Ordinary Person (talk) 13:32, 9 November 2010 (UTC)

Broken maths tags[edit]

Some of the formula tags in this article aren't working.Ordinary Person (talk) 13:51, 9 November 2010 (UTC)

Could yoy please point out where exactly? Salih (talk) 16:33, 9 November 2010 (UTC)

Incorrect terminal velocity for Baumgartner?[edit]

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 (talk) 01:53, 8 March 2017 (UTC)