Terminal Velocity depends on two things: surface area and speed. These are inversely proportionate.

If both these variables affect terminal velocity, why do parachutes slow you down? Initially you had a small surface area but a fast speed- with the parachute you have a larger surface area but lower speed. You have increased one variable but decreased the other. Therefore why do parachutes decrease speed?

  • $\begingroup$ Did you not answer this yourself: "a larger surface area but lower speed"? Terminal velocity and [maximum] speed are much the same thing. $\endgroup$ – Henry May 1 '12 at 7:25

You say:

Terminal Velocity depends on two things- surface area and speed

but I think you're getting slightly mixed up about the terminology. The drag (i.e. air resistance) depends on surface area and speed, but the terminal velocity is the speed and it just depends on the surface area (and air temperature, density, etc , etc that we'll assume is constant). You say;

with the parachute you have a larger surface area but lower speed

and this is quite correct but the speed is the terminal velocity so with the parachute you have a larger surface area but lower terminal velocity.

  • $\begingroup$ This answer is only correct when assuming all objects have exactly the same mass per surface area. A 1 meter sphere of iron and a 1 meter sphere of styrofoam are going to have very different terminal velocities though they have the same surface area. The parachute only works because it keeps your weight roughly the same (negligible increase), but increases your surface area by more than an order of magnitude. If it increased your weight by the same factor as your surface area, it would have no effect on your terminal velocity. $\endgroup$ – Shufflepants Apr 18 at 21:51

$(Drag) = (Coefficient)\times(Area)\times(Speed)^2 $

When $(Drag)=(Weight)$ then

$$ (Speed) = \sqrt{ \frac{ (Weight) } { (Coefficient)\times(Area) } } $$

So the larger the area the less the speed. What is the question again?


Terminal Velocity depends on two things- surface area and speed. These are inversely proportionate.

Terminal velocity depends on two things: Drag force and gravity. These very much are not inversely proportional to one another. Terminal velocity is reached when the drag force is equal but opposite to gravitational force.

To a bacterium, drag force is proportional to speed. Bacteria live in extremely low Rayleigh number regimes. We humans "live" in high Rayleigh number regimes. In those regimes (falling coins, flying aircraft, re-entering spacecraft, and parachutes), drag force is proportional to the square of velocity.

Note that I've written drag force rather than surface area. Surface area does have a marked influence on drag force. However, different objects with the same surface area can be subject to markedly different drag forces. The shape of the object also comes into play. A nicely shaped wing will offer much less air resistance than will an object such as a parachute even if the wing and parachute have the same surface area. The coefficient of drag distinguishes the wing from the parachute. The drag force on an object moving at some velocity $v$ is given by $F_d = \frac 1 2 \rho v^2 c_d A$, where $\rho$ is the density of air, $v$ is the velocity of the object, $A$ is the cross section area, and $c_d$ is the coefficient of drag.

By design, a good wing has a very low coefficient of drag and a low cross sectional surface area. By design, a good parachute has a very high coefficient of drag and a large cross sectional surface area. It's the combination of the two factors (coefficient of drag and surface area) that make terminal velocity so low for an object descending with a parachute.


The answer is that it decreases your net density. The person has a density of 998 Kg/m3. The parachute has a mass of 7Kg so that increases the Kg by 7 and the m3 by 0.2*0.3*0.3 = 0.05 cubed New density =1006Kg per 1.07 m3 = 958.09 Kg per meters cubed. Next we add the air in the parachute at 1.225 Kg/meters cubed The volume =1/2 x 4/3 pi r cubed approximately as it is parabolic actually d=11 ft = 3.3528 m = r=1.6764 m

volume = 39.4685 m cubed

So the mass has gone up by 48 Kg and the volume has gone up by 39.4685 m cubed So we arrive at 1045.04685 Kg per 40.5385 m cubed = 25,77 Kg/m3

So the parachute changes your net density from 998 Kg/m3 to 25,77 Kg/m3

This is compared to air at 1.225 Kg /m cubed at sea level or 0.018 Kg/m3 at 30,000 ft

The terminal velocity is determined from the ratio of the instantaneous density of the air and the net density of 25.77 Kg/m3 X the ratio of the viscosity of the person wearing a parachute/the air. The viscosity also has to be multiplied by a variable related to the Renolds Number to scale it up.

The answer then, without the parachute 125 mph

             with it 25.77/1.225 x 9.7126415 E-5/1.81206E-5 = 5.36 m/s

             5.36 m/s = 12 mph    assuming laminar flow and Renolds no =1 

When we are in water we are at our Natural density layer surrounded by a medium very similar to our density. Some people float on the water. Float meaning having the same natural density as the surface of water. Others are more dense and are stable 5m below the surface of the water. We are at this natural density as our origins were where we once lived, in the sea. So our density is very similar to sea water that has slightly higher density than pure water.

Once at our natural density level we feel no pull. So unless an object is not in it's natural layer it will not feel any gravity force. This gravity force is just the sorting force applied by the net inertia applied towards the centre of the Earth brought on by the differences of the inertia at the side of the Earth being further away from the sun and nearer to it as the Earth rotates. This creates a squashing inertia from the just over half of the world and a pulling inertia from the side nearest the sun. The plane that is the same as the Earths inertia, is an 3D arc shape of equal distance to the sun, this passes through the centre of the Earth. The net greater squashing inertia than the pulling inertia creates a net inwards inertia and as the Earth rotates on a tilted axis, this inertia is then directed at the centre of the Earth. The inertia is only converted into a force when there is a difference in density between materials. Imaging a pile of competing molecules all lined up with the centre of the Earth and free to move. The densest molecules will always have a positive density and a force will be created to move it around the next molecule until after many handshakes it finally reaches the centre of the Earth. All this turmoil has long sinced stabilised and only materials placed out of their natural density level by man usually are subject to a sorting force. If the material is blocked from moving freely by a more dense material in the way it carries on feeling the sorting force.

When it rains many tonnes of rain are dropped but soon it seeps down through the less dense earth then through the gaps in the rocks to underground streams that eventually reach rivers and finally the rain gets back to the rains natural density level, the sea.

We are drawn to go to the sea side for the same reason

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    $\begingroup$ I am much more dense than the average human and I prefer the mountains to the ocean. Explain that one. $\endgroup$ – Asher Jun 21 '15 at 22:39
  • $\begingroup$ Your density then must be more similar to air than most. This may be that your ancestral fish relatives climbed out of the sea much earlier and you have had more time to evolve from your natural density level of water to your evolutionary natural density level of air. We will all turn into baloons of hot air eventually. $\endgroup$ – Mike Kenyon Jun 21 '15 at 22:45
  • $\begingroup$ You do realize that evolution has no direct effect on an individual's density and that density is a physical, not metaphysical, quantity, right? Besides that, your beliefs about evolution are rather nonsensical. $\endgroup$ – Asher Jun 21 '15 at 23:22
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    $\begingroup$ This is just nonsense. $\endgroup$ – Kyle Kanos Jun 22 '15 at 0:02
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    $\begingroup$ For VLQ reviewers, I'll advance my opinion that this is a very bad answer, but it is an attempt to answer to question. $\endgroup$ – 299792458 Jun 22 '15 at 4:38

protected by Qmechanic Nov 15 '17 at 13:27

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