Aircraft Design Tutorial: Airspeeds Made Simple - Part 1 of 2

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  • เผยแพร่เมื่อ 3 ธ.ค. 2024

ความคิดเห็น • 21

  • @PatrickJWenzel
    @PatrickJWenzel 6 ปีที่แล้ว +1

    Incredibly concise, practical and simply explained. Perfect for people trying to understand compressibility effects with altitude.
    Thank you for going to the trouble of creating these two videos.

  • @soban8sky
    @soban8sky 7 ปีที่แล้ว +6

    Explained in a simple and graphic way. I wish you were my math and physics teacher.

  • @manu2790
    @manu2790 8 ปีที่แล้ว +3

    This is great explanation! I had to re vise these concepts for my private pilot! :)

    • @dr.gudmundssonaircraftdesign
      @dr.gudmundssonaircraftdesign  8 ปีที่แล้ว

      Hi Manu. Good to see you're still around. Congratulations with you private pilot's license.

  • @cesgar7994
    @cesgar7994 7 ปีที่แล้ว +2

    Is there an error at the 11:09 table? Headwind on true speed says 25knots (like the example) but the equivalent speed says 20. Is there an error? Or maybe I am not understanding something. 😢

    • @dr.gudmundssonaircraftdesign
      @dr.gudmundssonaircraftdesign  7 ปีที่แล้ว +3

      Hi Ces Gar. Thank you for your question.
      No, there is not an error. The center column contains true airspeed and the right column contains the corresponding equivalent airspeed, which is calculated using Veas = Vtas∙sqr(sigma), where sigma is the density ratio and "sqr" means square root. (I mean no disrespect by stating this if you already know this - I just do so to side with more rather than less clarity).
      Standard day density ratio for 15000 ft is 0.6292. Recall that the speedometer (50 knots) shows ground speed. We also know that the wind speed (25 knots) is true airspeed. Now imagine we're driving on a windless day at 50 knots. On that day the ground speed is equal to true airspeed. This helps one realize that the ground speed is related to the true airspeed from Vgs = Vtas + Vwind. The only difference is that on the windy day we have to fight a 25 knot headwind. Thus, an observer sitting outside on the car will experience 50+25 = 75 knot true airspeed. We can now use the above formula to calculate the corresponding equivalent airspeeds. Thus, for the 50 knots we get: (50)∙sqr(0.6262) ~ 40 knots (eas). For the wind contribution we get (25)∙sqr(0.6292) ~ 20 knots. Of course the symbol "~" means "approximately" and the numbers are rounded. Note that algebra allows us to sum it up as shown in the table.
      I hope this explanation helps Let me know if it doesn't.

    • @nkkkn8320
      @nkkkn8320 7 ปีที่แล้ว +1

      Great

  • @cristoballozano2143
    @cristoballozano2143 6 ปีที่แล้ว +1

    Thanks, great video... But im still confused with one thing. Are the KIAS and KEAS the same value without wind? Like TAS and GS?(without wind)? If i were at MSL i would have 40 KTAS and also in my ASI, at 15,000 ft i would need 50 KTAS to still have 40KIAS. In your example, those 40Kt are Equivalent Airspeed.

    • @dr.gudmundssonaircraftdesign
      @dr.gudmundssonaircraftdesign  6 ปีที่แล้ว +1

      If there is no error in the instrument system and no compressibility effects (low Mach), then KIAS and KEAS are the same value. Yes, in that case if you see 40 KIAS on your ASI at MSL, you would be doing 40 KEAS. Then, at 15000 ft if you see 40 KIAS on your ASI, your EAS would also be 40 KEAS and TAS would be 50 KTAS (no wind condition). The dynamic pressure associated with 40 KEAS is 5.42 lbf/ft² (psf). So, since the ASI is really a (dynamic) pressure sensor, anytime the dynamic pressure reaches 5.42 psf, the ASI will show 40 KIAS, no matter the altitude. So, in order to develop a dynamic pressure of 5.42 psf at, say, 15000 ft, the true airspeed must amount to 50 KTAS, even though the ASI will only show 40 KIAS. I hope this makes sense; I know it is a bit confusing. Best wishes!

    • @cristoballozano2143
      @cristoballozano2143 6 ปีที่แล้ว

      @@dr.gudmundssonaircraftdesign Got it, thank you very much!!

  • @ada6a92
    @ada6a92 7 ปีที่แล้ว +2

    Very nice videos, thank you.

  • @alfaisalwork504
    @alfaisalwork504 ปีที่แล้ว +1

    This is great. I wish your videos were translated into Arabic😔😔😔

    • @dr.gudmundssonaircraftdesign
      @dr.gudmundssonaircraftdesign  ปีที่แล้ว

      Yes, sorry about that. Thanks for watching, nevertheless. Best wishes.

    • @alfaisalwork504
      @alfaisalwork504 ปีที่แล้ว

      @@dr.gudmundssonaircraftdesign I've seen a free pdf copy of your book, have you published it?

  • @tonypeden8092
    @tonypeden8092 3 ปีที่แล้ว

    This is a complicated subject but if you're going to use a car's speedometer as the reference then the analogies should be formed around ground speed = true airspeed with no wind. Equivalent airspeed just confuses the subject.

  • @chafaon
    @chafaon 5 ปีที่แล้ว +1

    Great dtuff