# Equation for falling objects that includes terminal velocity

## Main Question or Discussion Point

if an objects terminal velocity in air is 18m/second, when what would be an equation that describes its velocity from t0?

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I am a technician, and have asked multiple engineers and new graduates, and with acceleration going to 0 when an object reaches it's terminal velocity seems to be more physics that they can handle.
V= A*T but A goes to 0 when V=18m/s, so V needs to nested in A, and I have no idea as to how to do that, if not an equation like f(v). i am also 57 years old, this is more of a fun thing

Ibix
You have a downward force on the object, which is gravity. This is ##-mg##. You have a force against the direction of movement that grows with speed, which is air resistance. This is some function ##f(v)##. Thus the net force is ##F=f(v)-mg##, which of course is equal to ##ma##. You reach terminal velocity when your velocity is such that this is zero.

To answer your question more specifically than this you need to know what ##f(v)## is. This depends on the shape and mass of the object and probably other stuff too, like what the surface is made of and the air density. There won't be an analytical answer to that question. I would expect you would need computational fluid dynamics to answer it.

yes, I fallowed that, so If I stick a cantaloupe out the window of the car and when I go 20m/s the angel is 45 degrees then the terminal velocity is 20m/s and the force of gravity is equal to the air resistance.

Filip Larsen
Gold Member
Terminal velocity is usually associated with vertical motion, and for that the equations of motion can be found in closed (analytic) form for both linear and quadratic drag. I know of a at least one thread from 2010 on this forum that describes it and you should also be able to find it elsewhere on the net. Whether or not you feel this will answer your question is another story, I guess.

DaveC426913
Gold Member
yes, I fallowed that, so If I stick a cantaloupe out the window of the car and when I go 20m/s the angel is 45 degrees then the terminal velocity is 20m/s and the force of gravity is equal to the air resistance.
No. Its terminal velocity is still 18m/s. Regardless of how fast or slow it starts at, that is the velocity that it will approach, due to air resistance.

Also, in the <2m distance from your hand to the ground, it will not reach its terminal velocity.

The cantaloupe is on a rope, from a pole sticking out of the window and if the terminal velocity is 18m/s, when the car is going 18m/s, the anel of the rope will be 45 degrees. Vt= G.
I get this:

I need to figure out how to put this in an Arduino. If i look at the slope in 10ths of seconds along the curve I should be able to Vt
I got to get some college kid to do the real calculus. I had Calc 3 was in 1998 :) the application is, if my airplane is at 500 feet and doing 40MHP what is the lead angle to hit an abandoned car with said cantaloupe. Physics is fun

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jbriggs444
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The cantaloupe is on a rope, from a pole sticking out of the window and if the terminal velocity is 18m/s, when the car is going 18m/s, the anel of the rope will be 45 degrees. Vt= G.
I get this:View attachment 238034
I need to figure out how to put this in an Arduino. If i look at the slope in 10ths of seconds along the curve I should be able to Vt
I got to get some college kid to do the real calculus. I had Calc 3 was in 1998 :) the application is, if my airplane is at 500 feet and doing 40MHP what is the lead angle to hit an abandoned car with said cantaloupe. Physics is fun
You are still up against the same basic problem. You have an unknown curve (the graph of drag versus velocity) for which you know only one parameter: f(18) = 1.

Without additional information, the problem you pose cannot be solved.

The web page you are looking at addresses that problem by assuming linear drag. So that, for instance f(v) = v/18. It seems to me that linear drag is a particularly simple case where the vertical and horizontal differential equations de-couple.

For unforced motion, the differential equation resolves to exponential decay. v(t) = v(0) * e-kt
For motion under a constant force, one simply adds the terminal velocity under that force to the unforced solution.

One can solve for vertical velocity as a function of time, integrate and invert to get the time to fall from a given altitude. One can solve for horizontal velocity as a function of time, integrate and evaluate at t = time to fall to get forward progress while falling.

Given x and y offsets, the target down-angle at which to drop is simple trigonometry.

The Norden bomb sight does the same thing with somewhat different inputs.

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Filip Larsen
Gold Member
The web page you are looking at addresses that problem by assuming linear drag.
Just a heads up, that the drag for on an object near terminal velocity in atmospheric air is going to be dominated by quadratic drag. See for instance [1] for description of how linear and quadratic drag depends on the Reynolds number.

[1] http://www.physics.emory.edu/faculty/weeks//journal/owen-ejp05.pdf

jbriggs444
Just a heads up, that the drag for on an object near terminal velocity in atmospheric air is going to be dominated by quadratic drag. See for instance [1] for description of how linear and quadratic drag depends on the Reynolds number.
For Reynolds number small that is not what this article says. It depends therefore on the density of object.

Filip Larsen
Gold Member
I am not sure why you say that.

The paper I linked to describes ways to model linear and quadratic drag, i.e. drag force varying linearly and quadraticly with respect to speed, with linear drag mostly being associated with low Reynold numbers (Re << 1) and quadratic drag with high Reynold numbers (Re > 1000). The goal of the paper is to then go on describing how to model drag when 1 < Re < 1000. In context of this thread, my point of linking to the paper was really just to say that if you stick a sphere-like object out the window of a car moving around 18 m/s (Re >> 1000) then drag clearly is in the quadratic regime.

It depends therefore on the density of object.
If by "it" you mean drag, then that is not correct since the drag force on an object do not depend on the density of that object. Object density do come into play for dynamical problems (like for instance a free falling object near terminal velocity) usually expressed via the so called ballistic coefficient.

hutchphd
I am not sure why you say that.

The paper I linked to describes ways to model linear and quadratic drag, i.e. drag force varying linearly and quadraticly with respect to speed, with linear drag mostly being associated with low Reynold numbers (Re << 1) and quadratic drag with high Reynold numbers (Re > 1000). The goal of the paper is to then go on describing how to model drag when 1 < Re < 1000. In context of this thread, my point of linking to the paper was really just to say that if you stick a sphere-like object out the window of a car moving around 18 m/s (Re >> 1000) then drag clearly is in the quadratic regime.

If by "it" you mean drag, then that is not correct since the drag force on an object do not depend on the density of that object. Object density do come into play for dynamical problems (like for instance a free falling object near terminal velocity) usually expressed via the so called ballistic coefficient.
My apologies. You were referring to the "cantaloupe on a rope" experiment where I was talking about the "falling object" which was the original question. I just wanted to clarify that for an arbitrary falling object the linear velocity drag regime can be appropriate. It is sometimes difficult to follow the thread ....

Filip Larsen
Gold Member
My apologies. You were referring to the "cantaloupe on a rope" experiment where I was talking about the "falling object" which was the original question.
Fair enough.

I just wanted to clarify that for an arbitrary falling object the linear velocity drag regime can be appropriate.
For free falling objects under one gravity in atmospheric air linear drag would perhaps be appropriate for dust flakes or similar microscopic objects, in which case the bouyancy force should also be considered as it becomes comparable to the drag force.

I still maintain that a quadratic drag model is most appropriate for cases involving air, gravity and objects you can see.

Fair enough.

For free falling objects under one gravity in atmospheric air linear drag would perhaps be appropriate for dust flakes or similar microscopic objects, in which case the bouyancy force should also be considered as it becomes comparable to the drag force.

I still maintain that a quadratic drag model is most appropriate for cases involving air, gravity and objects you can see.
Yes i now understand you are correct. Maybe very small insects and. for instance, Millikan oil drops but much larger goes too fast and any smaller the fluctuations take over. Thank you for taking the trouble to educate me!

jbriggs444
Homework Helper
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dust flakes or similar microscopic objects, in which case the bouyancy force should also be considered as it becomes comparable to the drag force.
I am a bit curious about how buoyancy enters in. One would expect small metallic flakes (for instance) to have the same density as larger metallic ingots. If buoyancy is negligible for one, it should be negligible for the other as well.

Are you, perhaps, considering "fluffy" dust tufts where the behavior is well approximated as a combination of the fibrous tuft itself plus the air that is effectively entrained within?

I am a bit curious about how buoyancy enters in. One would expect small metallic flakes (for instance) to have the same density as larger metallic ingots. If buoyancy is negligible for one, it should be negligible for the other as well.

Are you, perhaps, considering "fluffy" dust tufts where the behavior is well approximated as a combination of the fibrous tuft itself plus the air that is effectively entrained within?
But the terminal velocity scales with mass

jbriggs444
Homework Helper
2019 Award
But the terminal velocity scales with mass
But the the downforce from gravity and the buoyancy from air scale identically. If downforce from gravity dominates at one scale, it dominates at all scales.

Viscous resistance does not scale identically, of course. But viscous resistance and buoyancy are different effects.

But the the downforce from gravity and the buoyancy from air scale identically. If downforce from gravity dominates at one scale, it dominates at all scales.

Viscous resistance does not scale identically, of course. But viscous resistance and buoyancy are different effects.
But the drag force goes up slower (maybe like an area not a volume). We don't worry about the buoyancy of an F-15 fighter!

Filip Larsen
Gold Member
I am a bit curious about how buoyancy enters in. One would expect small metallic flakes (for instance) to have the same density as larger metallic ingots. If buoyancy is negligible for one, it should be negligible for the other as well.

Are you, perhaps, considering "fluffy" dust tufts where the behavior is well approximated as a combination of the fibrous tuft itself plus the air that is effectively entrained within?
I admit I have very little specific knowledge about the aerodynamics of dust flakes. I was thinking that the "effective density" of a flake, if compared to a sphere of same mass and characteristic size (e.g. diameter, projected area or similar), would be low, but I agree that this is not relevant for whether or not buoyancy is significant enough to be included. Thinking about it I would say that if someone said that the seemingly long "fall time" for dust has nothing do to with buoyancy but can alone be explained as a consequence of its terminal velocity being significant lower than normal circulation air speeds in a room (e.g. due to heating, people moving about, etc) then I would think that sounded quite plausible. I guess it would be easy to test, but I am also sure my wife would not approve if I start to conduct home experiments with falling dust, so I am unfortunately left speculating.

jbriggs444
Filip Larsen
Gold Member
But the drag force goes up slower (maybe like an area not a volume). We don't worry about the buoyancy of an F-15 fighter!
For the sanity of this discussion (when it involves cantaloupes and F-15 fighters, at least) I suggest you guys better forget I mentioned buoyancy. I mentioned it once and thought I got away with it.

jbriggs444 and hutchphd
For the sanity of this discussion (when it involves cantaloupes and F-15 fighters, at least) I suggest you guys better forget I mentioned buoyancy. I mentioned it once and thought I got away with it.
I remember doing the Millikan experiment (>40 years ago!) and using linear air drag with a buoyancy correction to ascertain the drop size by driving it up and down in the "telescope". So you are absolved of responsibility.

FactChecker
Gold Member
The drag force is proportional (proportional constant ##c##) to ##v^2## and it equals the gravitational force, ##g*mass## for ##v = 18m/second##. With that, you should be able to solve for ##c##. Use the resulting drag force, ##D = c*v^2##, in your equations for acceleration due to gravity (## A = g - D/mass ##) and integrate to get velocity.

jbriggs444
Homework Helper
2019 Award
I remember doing the Millikan experiment (>40 years ago!) and using linear air drag with a buoyancy correction to ascertain the drop size by driving it up and down in the "telescope". So you are absolved of responsibility.
Buoyancy correction should be around one part in one thousand for oil in air. It's an easy correction -- just subtract the density of air from the density of oil. See https://www.physics.uci.edu/~advanlab/millikan.pdf where the effective density of the oil, ##\rho'## is defined in that manner.

russ_watters
Mentor
I am a bit curious about how buoyancy enters in. One would expect small metallic flakes (for instance) to have the same density as larger metallic ingots. If buoyancy is negligible for one, it should be negligible for the other as well.
I deal with this for clean rooms, where larger particles settle out of the air faster than smaller ones. My understanding was that as the particles get smaller they behave more like gas molecules. I guess I never went deeper into what that means, but I guess that random molecular motion and turbulence become a larger contributor to a particle's movement as it gets smaller. This is why gasses don't stratify by density much in a room.

Or to say another way; I don't think it's that buoyancy goes down as particles get small, but rather air turbulence goes up to the point where we can't ignore it anymore.

Tom.G and jbriggs444