Is it possible to calculate max-q based on just upwards velocity instead of specially inputting a given altitude/air density? Based on this graph of Falcon 9 launches till MECO below "Air Pressure vs. Altitude", is it accurate to say that once the rocket goes through "the line of air pressure" is when max-q would occur?

I know the x axis isn't in distance km, but hopefully it help to get the question across.


Does passing through the air pressure "blue line" mean that the rocket will go through max-q as seen from the Falcon 9 graph below?

enter image description here

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    $\begingroup$ The value of dynamic pressure during an ascent is driven by increasing velocity and decreasing ambient pressure. It would be hard to calculate max q without the latter. $\endgroup$ Sep 15, 2018 at 23:09
  • $\begingroup$ What does "...once the rocket goes through 'the line of air pressure' is when max-q would occur" mean? $\endgroup$
    – uhoh
    Sep 15, 2018 at 23:25
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    $\begingroup$ With the calculation above, you need to enter the altitude the get the air pressure. Is there such calculation where you could determine max-q based on velocity through altitude per second? Of course the faster you go, the faster you reach max-q. So you must be able to say "by going this fast, you will reach max-q by this altitude" surly. $\endgroup$ Sep 15, 2018 at 23:26
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    $\begingroup$ @uhoh from the falcon 9 meco graph, it can be seen that the "Max-q" range is just before the blue line at around 11-15kms and passing through the blue line on the chart signifies max-q. $\endgroup$ Sep 15, 2018 at 23:40
  • $\begingroup$ Thanks for your comments and clarifications. Have a look at the answer I've posted so far and see if it addresses your question. $\endgroup$
    – uhoh
    Sep 16, 2018 at 1:00

1 Answer 1


I'm not sure I understand the question completely, but I'll work off of the comment

Of course the faster you go, the faster you reach max-q. So you must be able to say "by going this fast, you will reach max-q by this altitude" surely.

and take it to suggest that there may be a way to include altitude and therefore density implicitly in an expression rather than have it appear explicitly.

What does implicit mean? If my acceleration from rest towards a finish line at a distance $d$ is $a$, then using $x = \frac{1}{2} a t^2$ I can say my time will be $\sqrt{2d/a}$ without ever having talked about my speed. It's in there because I used calculus to integrate acceleration to get velocity, and to integrate velocity to get position $x$ in order to get the $x = \frac{1}{2} a t^2$ equation, but once I have that, velocity is inside there (implicit) even though it's not visible.

In a practical situation, probably the answer is no, because each launch has a different trajectory - some start turning sooner so that their vertical climb rate is lower, whereas others stay nearly vertical for a longer time.

But if we stick to a vertical launch for simplicity, and stick to the simple definition of dynamic pressure $q$ of

$$q = \frac{1}{2} \rho v^2$$

what then?

Let's say I can use a simple formula for the drag force:

$$F_D = -\frac{1}{2} \rho C_D A v^2$$

where the coefficient of drag $C_D$ is a constant. The problem of course is that it is not constant and varies a lot with both velocity and density as you go supersonic, and that behavior, for a realistic rocket, does not come from a simple formula but instead from careful measurements or very complex numerical simulations on computers. That's another source of "no".

Here's an example from @RussellBorogove's answer:

Saturn-V drag coefficient archived from braeunig.us/apollo/saturnV

Also, see this answer for more reading.

Also, for gravity (I almost forgot!):

$$a_{Grav} = -GM_E/(R_E+x)^2$$

where GM_E is the standard gravitational parameter for Earth and $R_E$ the Earth's radius, and of course $a_{Grav}$ is the acceleration due to gravity (since mass drops out later).

But what if we were launching on a crazy planet where max-q happened at fairly low velocity and we could treat drag with a simple formula?

The acceleration $a$ is then given by

$$a = \frac{F_{tot}}{m(t)} = \frac{F_{Thrust} + F_{Drag}}{m(t)} + a_{Grav}$$

and with a fixed flow of propellant $\frac{dm}{dt} = \dot{m}$ we can say

$$m(t) = m_0 - \dot{m}t$$


$$F_{Thrust}=\dot{m} v_{ex}$$

where $v_{ex}$ is the exhaust velocity. That gives

$$a = \frac{\dot{m} v_{ex} + F_{Drag}}{m_0 - \dot{m}t} + a_{Grav}$$


$$a = \frac{\dot{m} v_{ex} - \frac{1}{2} \rho C_D A v^2}{m_0 - \dot{m}t} - \frac{GM_E}{(R_E+x)^2}$$

I still need to know the density for the definition of $q$, and we can use a simple scale height approximation which will look a bit like your plots of pressure versus altitude. Assuming the atmosphere's temperature is constant (which it is not) we can say density and pressure are always proportional. Then

$$\rho(x) = \rho_0 \exp(-x/h_{scale})$$

Plugging that back into the equation for acceleration, you get

$$a(t) = \frac{\dot{m} v_{ex} - \frac{1}{2} \rho_0 \exp(-x(t)/h_{scale}) C_D A v^2(t)}{m_0 - \dot{m}t} - \frac{GM_E}{(R_E+x(t))^2}$$

That's an equation that has all three of these: height, velocity, and acceleration ($x$, $v$ and $a$) and so solving for $v(t)$ and $x(t)$ in order to get

$$q(t)=\frac{1}{2} \rho(t) v^2(t) = \frac{1}{2} \rho_0 \exp(-x(t)/h_{scale}) v^2(t) $$

and then solving for


in order to find the maximum is going to be quite the challenge!

I'm not aware of an analytical (simple equation) solution for $v(t)$ and $x(t)$ even in this highly simplified case, so I am going to go out on a limb and say "no" there isn't a way to get time to max-q without really solving the whole problem numerically in a computer, getting the altitude and velocity from the trajectory, then plugging the numerical results back into the equation for $q$ to get its maximum.

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    $\begingroup$ @UndefinedUsername I'm still not sure what "..exceeds the 'blue air pressure line'...'" means. Max-q happens when q is the largest. I'm not sure how exceeding a line comes into play. $\endgroup$
    – uhoh
    Sep 16, 2018 at 2:09
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    $\begingroup$ @UndefinedUsername yes indeed, I believe so. The definition of $q$ is simple, $\frac{1}{2} \rho v^2$ but there's no simple way to get $\rho$ or $v$ without a complicated simulation. Unless someone makes a convincing argument otherwise, I think that it is safe to assume that it's just the same kind of number crunching that they use to plan and program the launch, and to design the rocket in the first place and put the right amount of fuel into it for a given payload mass. $\endgroup$
    – uhoh
    Sep 16, 2018 at 2:45
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    $\begingroup$ So in basic, without a complicated simulation, the computation of max-q given a calculation of thrust is roughly infeasible as can easily be seen from your equation. Anyway, thank you for trying. It is to be expect as some kind of system. Their press kits are executed on the dot hours or even possibly days before the launch.It just happens that half of the formula requires an unknown altitude of a rocket configuration without simulation. $\endgroup$ Sep 16, 2018 at 3:01
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    $\begingroup$ @UndefinedUsername I think the nominal mission plan is calculated well ahead of time, along with many alternates, since the launch window can be very wide in some cases (hours) even though SpaceX has these one-second "instantaneous" launch windows. It's an interesting topic and I would recommend you ask more questions about it! (Check for existing questions first) $\endgroup$
    – uhoh
    Sep 16, 2018 at 3:02
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    $\begingroup$ About that 'blue line' ; 'crossing the line' doesn't, as far as I can see, have any direct significance to the parameters uhoh deliniated - it is simply an artifact of the graph that the line runs close to the calculated 'points' of Max-Q. The line indicates air pressure (either measured or predicted) around the vessel as it ascends and proceeds downrange; it follows naturally that the 'knee' occurs just after Max-Q for most launches, since as air resistance recedes acceleration naturally increases for constant thrust, hence velocity will increase and air pressure will drop more rapidly. $\endgroup$ Dec 15, 2018 at 4:55

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