On a conceptual level, what's happening in this equation?

[cmkluza]

I'm trying to read up on some traffic flow models and I've encountered the following equation (source [pdf]):

Variables are $t =$ time, $x =$ space variable along road, $\rho (t, x) =$ density of cars, $v(\rho ) =$ velocity, $Q = \rho \times v$

Following graphic is used to show situation:

If total number of vehicles is conserved:
$$\frac{d}{dt} \int^b_a \rho (t, x) dx = Q_a - Q_b$$
$$\frac{d}{dt} \int^b_a \rho (t, x) dx = \rho (t,a)v(\rho (t,a)) - \rho (t,b)v(\rho (t,b))$$
What are these equations actually showing, namely the first bit (derivative of integral of density)? I'm a bit confused since I'm not too familiar with the math behind it. Ignoring numbers and math right now, what are the physical interactions described by this? That might not be worded to clearly, so what does the integral of the density with respect to distance show? What, then, does the derivative with respect to time show?

Sorry if this is asking a lot, but any help understanding this will be appreciated!

 

[russ_watters]

That looks to me like a mathematical description of the accordion effect in traffic: one person in a line of traffic taps their brakes and before you know it, you have a backwards propagating wave of people slamming on their brakes and barely avoiding rear-ender accidents.
https://en.wikipedia.org/wiki/Accordion_effect

 

[256bits]

What are these equations actually showing, namely the first bit (derivative of integral of density)? I'm a bit confused since I'm not too familiar with the math behind it. Ignoring numbers and math right now, what are the physical interactions described by this? That might not be worded to clearly, so what does the integral of the density with respect to distance show? What, then, does the derivative with respect to time show?

Hi
Well to start with, the density in linear density, example: cars/meter
Q is the flow rate of cars in cars/sec . that easily can be seen from Q = ρ x v, Example : Q = cars/meter x meters/sec ==cars/sec.

Density is some function of time and position.
For any section ab, the integral of the density function yields the value of the density between ab.
The derivative yields the rate of density change per time.

That last part is not worded quite correctly, but maybe you (or someone else) can correct it.

cherios

 

[cmkluza]

That looks to me like a mathematical description of the accordion effect in traffic: one person in a line of traffic taps their brakes and before you know it, you have a backwards propagating wave of people slamming on their brakes and barely avoiding rear-ender accidents.
https://en.wikipedia.org/wiki/Accordion_effect

Ah, perfect, that's what I'm hoping to develop in my investigation! Thanks for the link, I keep learning new things about traffic flow; it's a much larger subject area than I'd originally thought.

Hi
Well to start with, the density in linear density, example: cars/meter
Q is the flow rate of cars in cars/sec . that easily can be seen from Q = ρ x v, Example : Q = cars/meter x meters/sec ==cars/sec.

Density is some function of time and position.
For any section ab, the integral of the density function yields the value of the density between ab.
The derivative yields the rate of density change per time.

That last part is not worded quite correctly, but maybe you (or someone else) can correct it.

cherios

As far as I know, that last part is worded correctly; at least it's helped me gain a better understanding of this equation. Thanks for your input!

 

[cmkluza]

@russ_watters and @256bits
I hate to bother you guys with too many questions, but could you explain why exactly the change in flux is equivalent to the derivative of the integral of density? I believe I understand what the derivative and integral term is saying for the most part now, it's showing the rate of density change per time, but how does this relate to change in flux? Change in density per time should be
$$\frac{\frac{cars}{distance}}{time}$$
while change in flux should just be
$$\frac{cars}{time}$$

Do you guys know how/why they are equating the two?

Edit: Is it possible that the initial integral of density yielded only the number of vehicles within the segment over which it was integrated?

 

[Blibbler]

Try drawing it! Get a pen and a piece of paper. Make different scenarios for traffic flow rate, density etc. and see empirically how the equations describe what you are drawing.

All equations ultimately describe systems which you can visualise, so by turning this on its head you can make a visual representation of the traffic flow and gain an intuitive understanding of the equations. There's really nothing abstract about it at all.

 

[256bits]

@russ_watters and @256bitsEdit: Is it possible that the initial integral of density yielded only the number of vehicles within the segment over which it was integrated?

Yes that is what it means. Hence the needed correction I mentioned, which you have now solved.
congrats. Give yourself credit.

 

[M Quack]

I'm trying to read up on some traffic flow models and I've encountered the following equation (source [pdf]):

If total number of vehicles is conserved:
$$\frac{d}{dt} \int^b_a \rho (t, x) dx = Q_a - Q_b$$
$$\frac{d}{dt} \int^b_a \rho (t, x) dx = \rho (t,a)v(\rho (t,a)) - \rho (t,b)v(\rho (t,b))$$

The integral from point A to point B over the density of cars is just the number of cars between these two points.

The time derivative of the number of cars is the difference between the number of cars going in on one side and coming out on the other side.

In all of this the traffic is simulated as a continuous medium. Rather than having a car at one point, then nothing for a few meters, and then another car, there is a continuous density of cars, say 0.1 cars/meter or 1 car per 10 meters. This makes the simulation a lot easier, and it makes no difference to the result if you are only interested in trends of what happens if there are many cars.

The number of cars passing through a point A (Q(x=A, t) is then the density of cars at that point and at the time t when you are looking, rho(x=A,t), times the velocity of cars at that point and time, v(x=A, t).

If somewhere between point A and point B Godzilla was snacking on cars, then the equation would no longer be valid.