Rocket Science: Stability - Center of Pressure Mikes Inventions

Поділитися
Вставка
  • Опубліковано 4 лют 2015
  • mikesinventions.etsy.com mikesinventions.etsy.com
    This is basic rocket science. I explain a bit about stability & show an easy way to calculate the rocket's center of pressure

КОМЕНТАРІ • 28

  • @Danimal-3D
    @Danimal-3D 6 років тому

    Great explanation -- I definitely appreciated that you touched on alternate fin design and how that impacts the calculations.

  • @aravindshreyas3571
    @aravindshreyas3571 8 років тому

    brilliant explaination!!
    thumbs up!

  • @stefanturcu6547
    @stefanturcu6547 3 роки тому

    Thank you for the video, Mike! Realy helpful!

  • @harshitbalaji2892
    @harshitbalaji2892 5 років тому

    Thank you sir for the explanation

  • @JaraSchatz
    @JaraSchatz 7 місяців тому

    Love it!

  • @wadihhelou3048
    @wadihhelou3048 3 роки тому

    thank u very much very useful explenation

  • @raphaelboll6096
    @raphaelboll6096 6 років тому +2

    Thumbs up Dr. Sheldon Cooper

  • @ahmadwardak836
    @ahmadwardak836 8 років тому

    Tanks so mache for your video

  • @nuggetthepenguin473
    @nuggetthepenguin473 3 роки тому

    But shouldn't u consider the pressure differences along the surface. Or can u just simplify it for the model rocket case by making pressure constant?

  • @vaibhavkumbhar3524
    @vaibhavkumbhar3524 3 роки тому

    Thank you very much.....

  • @adonikam1
    @adonikam1 3 роки тому

    I'm lost I only passed math for daily living!! Lol. For real thanks and great video!!!

  • @Rainal0704
    @Rainal0704 8 років тому +1

    fantastic video! I'm honestly surprised I cant find this solution anywhere else online. everything I've read so far said something about summing up the pressures and do it like your regular CM problem but none mention what to do when you don't know the pressure distribution. one question tho, is there a proof to show that M/A gets you CP?

    • @MikesInventions
      @MikesInventions  8 років тому +2

      +Rainal0704 Thanks! I'm glad it helped. As for the proof... the AIAA education series has tons of books on this. I also found something on a site run by NASA that says the same thing. When you have an area times a distance, you get a cubed dimension, and when you divide that by a squared dimension, you're left with a single number, which is your CP distance. Here's the NASA thing. I browsed around the site, and it explains a lot of the basics. Very nice site. exploration.grc.nasa.gov/education/rocket/rktcp.html

    • @Rainal0704
      @Rainal0704 8 років тому

      +Mike Thompson Thanks for the link, I'll read up on that! My initial guess is that it might have something to do with how force and pressure are related and in this case moment and area both contain the force element(which is cancelled by dividing) therefore allowing the calculation of cp to be done without knowing the force. anyways, thank you!

  • @egeres14
    @egeres14 7 років тому

    Nice video : )

  • @slap_my_hand
    @slap_my_hand 6 років тому

    I'm thinking abut building a remote controlled model rocket powered by an electric ducted fan. It's going to have air inlets on the sides, how does this affect the center of pressure?

    • @MikesInventions
      @MikesInventions  6 років тому +1

      That sounds interesting... So not a "rocket" technically, but it'll be cheaper to fly I bet. If the inlets protrude from the sides then you'll account for them just like you would a fin. If it's an internal NACA style inlet, then it doesn't affect anything. Either way, just project the area of the rocket and find the center. Have Fun!

  • @Workerbee-zy5nx
    @Workerbee-zy5nx 5 місяців тому

    Got it.

  • @aksh_x_
    @aksh_x_ 3 роки тому +1

    Thanks a lot!
    Also I was the 400th like hehe

    • @MikesInventions
      @MikesInventions  3 роки тому +1

      Well thank you!

    • @aksh_x_
      @aksh_x_ 3 роки тому

      @@MikesInventions Can you make a video on how to control a rocket's roll, pitch and yaw using fins? Like which direction will the rocket go if fins rotate to a certain direction at an angle and stuff. Would be great!

    • @MikesInventions
      @MikesInventions  3 роки тому +1

      @@aksh_x_ I'm not sure I understand exactly what you're wanting to know. I'm not likely to make a rocket with controllable fins anytime soon, and I'd certainly never make a rocket with fins that weren't pointed straight. Could you explain in a bit more detail what you'd like to see? Perhaps I can get it done.

    • @aksh_x_
      @aksh_x_ 3 роки тому

      @@MikesInventions ua-cam.com/video/qnUfsLCa8h8/v-deo.html Check this out, somewhat like this. Instead of gimbaling the SRM(as done in advance rockets), fins would work for making the rocket go on a certain path or to make it stable. In the video they have added small fins on the top for that purpose but is it possible to do it using the bottom fins only? And if yes I would love to get a detailed explanation of it!

    • @aksh_x_
      @aksh_x_ 3 роки тому

      @@MikesInventions Also you can just explain the physics and aerodynamic properties behind it, would be a great help!

  • @mohamedelaminebellebna7965
    @mohamedelaminebellebna7965 4 роки тому

    How we can make difference between 2 rocket which have 2 fins and 4 fins.they have the same center with this method

    • @MikesInventions
      @MikesInventions  4 роки тому +1

      Hey Mohamed! There is no difference, however you'd never make a rocket with only 2 fins. 3 is the minimum required for stability. If you think about a rocket with 2 coplanar fins, it could rotate in the plane of the fins, and that's not good. If you were to have fins of different sizes for some reason, you'd need to calculate the stability of the rocket in pitch and yaw, because it would have different characteristics in each plane. Now, you could make a rocket with only 2 fins if they weren't coplanar, but why do that? The math for stability is much more difficult. a 4 fin rocket with equal fins will have identical stability in pitch and yaw, that's why you only need to use the projected area, which is the same as just 2 fins. Hope that helps. Thanks for watching!

    • @mohamedelaminebellebna7965
      @mohamedelaminebellebna7965 4 роки тому

      @@MikesInventions it helps a lot thanks for this detailed explanation sir.