VD question

  1. Hello,

    I'm not an aerospace engineering student but I'm interested in aviation and I had a question.

    My friend and I got into a debate last night about VD. I believe he said that meant vertical descent. Anyway, he said that if an aircraft has a VD limit = 420 KCAS to 17,000ft/.91M that means it can fly up to that speed from 0ft-17,000ft. Is he correct?

    Also, I know what 420 knots is about 480 mph but what does .91M mean?

  2. jcsd
  3. I'm not 100% sure about the terminology of VD but http://www.gps.tc.faa.gov/glossary.html says it's the design dive speed. I know that you do not want to exceed the Vne which stands for velocity (never exceed) because exceeding that means that the airplane's frame could become overstressed and fail or other serious failures (flutter of control surfaces, inability to maneuver out of the dive) could occur.

    The only thing that makes sense with the .91M figure is .91 Mach, but that doesn't seem right because .91 mach is .91 times the speed of sound. The speed of sound at that altitude will be around 700mph, and 480mph is only around .7 mach...so I think .91M might mean something else.

    Maybe there's a private pilot here that can answer?
  4. Greg, Thank you for the quick response. The data for the plane in question is:

    Airspeed Limits:
    VD = 420 KCAS to 17,854 ft/.91M above 23,000 ft, linear variation between these points.

    Its for a boeing 767-200. I always thought large airplanes like that could fly at those speeds at high altitudes (cruising speeds). Wouldn't the air be too thick at low altitudes (i.e. <1200ft) for it to fly that fast? Wouldn't the engines stall?
  5. FredGarvin

    FredGarvin 5,087
    Science Advisor

    The .91M is the local Mach number. Even at 17,000 ft, 480 kts (I don't really care about calibrated vs. indicated airspeeds etc). Greg is correct that the Mach number doesn't jive in that context. Even at roughly 17,000 ft, the local speed of sound for standard atmosphere is around 715 mph. The way I interpret your restatement is that anytime above 23,000 ft the limit is .91M. The .91M will be different speeds at different altitudes. Anything below 17,854 you are limited to 420 kts. The linear variation means that between 17,854 and 23,000 ft, you'll have to work the numbers to work out the function that will give you the line you need to follow. The lower point will be the 420 kts number. The upper number will be:

    23,000 ft, speed of sound ~ 698 mph = 607 kts. Draw a line between the two and plot it vs. altitude going between 17854 and 23000.

    As far as the engine goes, it will run anywhere you want it to. However, flying that low and that fast creates a lot of drag and thus your fuel consumption goes through the roof. The higher altitude provides low drag and the engine produces less thrust, and thus less fuel is consumed. Plus the temperature is lower so you get a bit of a pick up in inlet air density as well.

    As far as the limits and why they are what they are, that will have to be explained by the airframer. I can tell you it will be a compromise of structural, aerodynamic and engine inlet restrictions.
  6. Thanks for the clarification on the subject of the linear variation. I'm still kind of confused on the rest though. Where does the bypass ratio come into play? Also, are aircraft designed to far exceed their design specs? Do manufacturers design the aircraft with huge margins? Could a 767 realistically exceed that airspeed limit by maybe 15%-25%?

    Also, could someone maybe explain or link me to an online source that talks about thin shell buckling. I'm not an engineering student so I'm not familiar with advanced physics but I've taken many different into to physics classes.

    Thanks again.
  7. FredGarvin

    FredGarvin 5,087
    Science Advisor

    The bypass ratio has really nothing to do with this. That is simply the ratio of the fan diameter and the core diameter. Well, it does, but not directly. It is a performance parameter but you won't hear someone say they can't descend past XXX rate because we have an 8:1 BPR.

    Civilian aircraft are designed with safety factors built in. They're not as large as you would expect but they are there. I wouldn't call them huge. They aren't huge because that usually translates into more weight or cost.

    Shell buckling is a kind of generic term. A shell is a structural element. How a shell feature buckles depends on its geometry, loading and boundary conditions. Shells are usually used to model thin walled vessels and things like aircraft skins. Is there something in particular you're looking for?

  8. Right.
    1. But wouldn't the BPR be important here because its important as far as thrust goes? Is there an equation or a concept that I can use to figure out the minimum thrust needed to boost that plane to a certain speed at a certain altitude? I'm sure it would include inlet air density.

    2. You say the margins aren't huge, but in your professional opinion do you think that they would be less that 20%?

    3. Thank you for the site on thin shell buckling. It really helped me to understand the concept better.
  9. I looked around for factor of safety information, and http://cat.inist.fr/?aModele=afficheN&cpsidt=4913559 mentions that the standard factor of safety for aircraft is around 1.5, meaning the structure is designed to be 50% stronger than it theoretically needs to be. However, the forces on an airframe will most likely be due to the difference in the square of velocity, not a linear relationship, so increasing velocity past design can make things go bad, quickly.

    Bypass ratio, from what I remember of my propulsion class, mainly determines how efficient an engine will be for a given flight regime. High bypass ratios are good for subsonic transport aircraft flying at higher altitudes, in general, whereas low bypass ratios are good for high flying, supersonic aircraft until you get to the point that ramjets become more efficient. For example, the SR-71 functions as a ramjet above certain speeds (around Mach 3).

    Determining the thrust to get an airplane to a certain altitude will depend on the air pressure, drag coefficient, and lift coefficient, along with the velocity you want.

    For a case where you know the C_L (coefficient of lift, is dependent on velocity), and the C_D (coefficient of drag, also dependent on velocity), you can set the lift force equal to the weight of the aircraft, and the drag equal to the thrust. And then, you can say:


    where rho is the air density (at altitude) and v is the velocity, and A is the frontal area of the aircraft. These equations will limit the operating regime of the aircraft. It doesn't matter if you have enough thrust to fly around at 60,000 feet if you don't have the lift to keep you in the air.
  10. Ok so I found some of the variables:

    rho=1.1887 kg/m^3
    wing area + fuselage cross sectional area = 303.7 m^2 = A

    To get C_D I used the calculator from the university of washington. Its the first link when you google "drag coefficient calculator".

    The variables I used in the calculator are: 700ft cruise altitude and 575mph cruise speed.

    Is that a correct C_D or is there a better way to find it?

    With that information, I calculated the thrust:

    thrust = .024 * (.5 * 1.1887 kg/m^3) * (257.8 m/s)^2 * 303.3m^2
    = 287535.68 Nm

    Now, the engines on the aircraft have a static thrust of 75925.8 Nm. Does that mean that this aircraft can't produce the amount of thrust needed to propel it to a speed of 575mph at this altitude?
  11. I would say yes, as flying at that speed at an altitude of only 700ft (dense, sea level air) requires a lot more thrust when the pressure is a fifth or tenth of that at altitude.

    What I'm curious about though is your units, you have thrust written as Nm when it should just simply be N. Nm is an energy measurement. If you came up with that unit from your calcs, you just missed the m.


    Now, this same plane might be able to get to that speed if it was at altitude and went into a dive, but depending on the vehicle it might not be able to structurally take that speed at that altitude.

    That thrust seems about the same as a single engine jet fighter, but that area seems to be on the large side (an f-16 has a wing area of 28m^2, which might be close to what you are calculating)
  12. I thought Nm was a weird way to put it. When I looked at the static thrust info for the engines it gave it me in units of ft*lbs, which I thought would be Nm. If you go on wikipedia and look up the specs for the Pratt and Whitney JT9D's it will tell you. I can't post urls until 15 posts.

    Anyway, do you think that is a correct value for C_D? I want to present this information to my friend so I don't want to look more incompetent in aeronautics than I already am.

    Yeah the wing area is for the Boeing 767-200.

    How would I go about figuring out C_L?
  13. Oh, haha, it lists lbf as the thrust measurements. That should be read pounds force, as opposed to pounds mass. English units can be confusing that way. The way I calculate mass in english is either in Slugs or Blobs, because the whole lbm thing confuses me.

    That sounds reasonable for C_D, but my specialty isn't airplanes. It seems around the right order of magnitude, anyway.

    What you can do for C_L is back calculate it by inputting your aircraft's weight, velocity, air density, frontal area, and solving for C_L. Then you could look up lift coefficients for the airfoil of your choice (will depend on angle of attack). Actually, you could really just simplify this process by using the approximation for thin airfoil theory and say that C_L=2*pi*alpha where alpha is the angle of attack in radians (degrees*pi/180 incase you don't remember). This should be good enough as long as your angle of attack is under 10 degrees. From this you can calculate the angle of attack necessary for level flight.

    In reality, most airfoils for heavy aircraft like that have enough camber in the wing so that they have a positive C_L at 0 degree angle of attack (alpha). Unless your wing is a delta though, you really can't go beyond 10 degrees without nonlinearity/stalling issues.

    And I doubt a 767's airframe and controls will handle that speed very well at sea level, but I could be wrong, haha
  14. FredGarvin

    FredGarvin 5,087
    Science Advisor

    You're really not staying on track here. First you were talking about surges at low altitudes which is what I replied to. Now you are talking performance calcs. Yes of course the BPR and the fan are important when looking at the engine cycle and the overall performance. Honestly, the calculation you are trying to get to here is a very complicated one. When we do performance calculations like this,we use what we call engine decks or models. They take into account a myriad of variables. When an airframer selects our engine to power it, we use our cycle deck to help design and answer questions just like that. It is definitely NOT a back of the envelope type of calculation. Here are some of the variables you would have to consider: Temperature, humidity, weight, altitude, fuel type, angle of attack, etc...You could probably fudge some numbers together, but it will not have any kind of real accuracy.

    1.5-2.0 are what I ams used to seeing. It does depend on the criticality of the component/system and the target weights and costs for that component. It is always a trade off and just one of thousands of engineering decisions that need to be made.

    I hope it helps. It's out of my area so I can't really give you much more than search results.
  15. Thanks a lot for all of your help guys. Your assistance is much appreciated.
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