Lift coefficient of rocket at vertical flight

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Hi,

I am looking into aerodynamic parameters for rockets (ogive nose, cylindrical body of different diameters, four fins at bottom) and have a question about the lift coefficient.

If a rocket is launched vertically, with zero incidence angle and wind, would that result in a very small lift coefficient (close to zero) due to its symmetrical shape? Unlike a camber airfoil, a rocket shape would not create a low pressure area vs a high pressure area, thus not generating lift? Is this a correct assumption?

Thanks in advance!
 

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  • #2
BvU
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Hello krihamm, :welcome:

Yes.
 
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Thank your for your reply!

Follow-up question: If I want to calculate the derivative of the lift coefficient with respect to incidence angle, and then plot the derivative as a function of speed - how do I proceed? Just calculate the lift coefficient for two arbitrary angles at a fixed speed, then calculate the derivative of those two calculations, and then do the same calculations for different speeds?
Hello krihamm, :welcome:

Yes.
 
  • #4
BvU
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Rather a short reply, I admit :smile:

Follow up: clarification requested: you want the lift coefficient for a rocket taking off (probably starting vertically and then gradually tilting ? ( [edit] the right term might be: yawing :rolleyes:)
Meaning it has a considerable vertical speed ?
In such a case the context for the conventional lift coefficient expression probably does not apply at all ! The velocity of the trajectory is almost completely aligned with the axis of the rocket.

[edit2] Aren't you more interested in the drag coefficient ?
Some nice formulas here or here

(because I'm not a rocket scientist, I simply googled 'ballistic missile drag coefficient' -- did you ?)
 
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Rather a short reply, I admit :smile:

Follow up: clarification requested: you want the lift coefficient for a rocket taking off (probably starting vertically and then gradually tilting ? ( [edit] the right term might be: yawing :rolleyes:)
Meaning it has a considerable vertical speed ?
In such a case the context for the conventional lift coefficient expression probably does not apply at all ! The velocity of the trajectory is almost completely aligned with the axis of the rocket.
Correct and correct.

I am running simulations on such a rocket that utilizes high speed numerics, so assuming the estimations of the drag coefficients are correct, would the method mentioned above be correct to get the derivative of the lift coefficient with respect to incidence angle as a function of speed?

Thank you!
 
  • #6
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Isn't the fun of this lift coefficient expression that it's somewhat independent of speed ?
And your simulations already account for the effect, so I wouldn't use arbitrary angles, but very small deviations from the calculated angle to get partial derivatives wrt incidence angle. Other than that, I think it's worth trying. But try to find something in the litterature too.

[edit] I am thinking of instabilities (you too?) and I wonder if they can be treated/found with this apporach.
 
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Isn't the fun of this lift coefficient expression that it's somewhat independent of speed ?
And your simulations already account for the effect, so I wouldn't use arbitrary angles, but very small deviations from the calculated angle to get partial derivatives wrt incidence angle. Other than that, I think it's worth trying. But try to find something in the litterature too.

[edit] I am thinking of instabilities (you too?) and I wonder if they can be treated/found with this apporach.
Independent of speed in what way? The speed squared is included in the denominator, so surely the lift coefficient is heavily dependent on speed?
 
  • #8
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No. The lift force is heavily dependent on speed, namely pretty close to quadratically. And also on density. These two are extracted and what's left is the ##c_L##.

It is largely determined by shape aspects, and there will be a lot less dependence on speed
 
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  • #9
A.T.
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The speed squared is included in the denominator, so surely the lift coefficient is heavily dependent on speed?
When apples cost 2$ / 1 apple, is the price per apple dependent on the amount of apples you buy?
 
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No. The lift force is heavily dependent on speed, namely pretty close to quadratically. And also on density. These two are extracted and what's left is the ##c_L##.

It is largely determined by shape aspects, and there will be a lot less dependence on speed
So the variations of the lift coefficient vs Mach number a rocket experiences is only due to the conventional lift coefficient expression not being valid for supersonic speeds (as you mentioned above)?

https://history.nasa.gov/SP-468/ch5-2.htm
 
  • #11
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as you mentioned above
What did I mention ? I don't recall mentioning that :rolleyes:

The dip in the picture in your NASA link when approaching Mach 0.7 will have to do with the shock wave drastically changeing the pressure dirstribution.

My dependence on speed link was intended to illustrate the lesser dependence of ##c_D## on Re.
 
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I was referring to when you said that the conventional drag lift expression does not apply for high speeds, the rest was my attempt at conclusions :biggrin:

Thank you for your help!
 
  • #13
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Still think there's a communication mishap
when you said that the conventional drag lift expression does not apply for high speeds
In post #4 I wrote
Meaning it has a considerable vertical speed ?
whereas the context of the lift coefficient formula is about a horizontal speed (probably meaning the vertical speed is so small it can be ignored ?)
 
  • #14
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What did I mention ? I don't recall mentioning that :rolleyes:

The dip in the picture in your NASA link when approaching Mach 0.7 will have to do with the shock wave drastically changeing the pressure dirstribution.

My dependence on speed link was intended to illustrate the lesser dependence of ##c_D## on Re.
Still think there's a communication mishap


In post #4 I wrote
whereas the context of the lift coefficient formula is about a horizontal speed (probably meaning the vertical speed is so small it can be ignored ?)
Yeah, that's what I meant. My bad for not clarifying horizontal vs vertical speed.
 
  • #15
boneh3ad
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I am really, truly confused about why we are talking about a lift coefficient on a rocket traveling vertically (or perhaps more correctly, traveling with its axis parallel to the travel/flow direction). What am I missing here? Are you trying to include the effects of the "lift" force normal to the flow direction for when the angle of incidence deviates slightly from perfectly zero? I'd think the overall lift would still be very, very close to zero and the moment coefficient would be much more important there.
 

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