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Foam can't hurt the shuttle?

  1. Aug 25, 2015 #1

    HCD

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    1. The problem statement, all variables and given/known data

    Problem statement in attached file.

    2. Relevant equations

    Drag force as a function of velocity for a body immersed in a fluid:
    $$ \boldsymbol{F}_D = \frac{1}{2} \rho \boldsymbol{v}^2 C_D A, $$
    where ## \boldsymbol{F}_D ## is the drag force, ## \rho ## is the density of the fluid, ## \boldsymbol{v} ## is the speed of the object relative to the fluid, ## C_D ## is the cross sectional area of the body and ## A ## is the drag coefficient.

    3. The attempt at a solution

    I'm trying to model the forces acting on the foam after it came loose from the shuttle. From the problem statement I'm assuming that the forces that should be considered are gravity, the drag force and whatever force is accelerating the foam in the air flow. That is
    $$ \boldsymbol{F} = m \boldsymbol{g} + m \boldsymbol{a}_A - \frac{1}{2} \rho \boldsymbol{v}^2 C_D A, $$
    but I'm not sure how to model the acceleration due to the air flow. I'm guessing that it has something to do with the velocity of the shuttle?
     

    Attached Files:

  2. jcsd
  3. Aug 25, 2015 #2

    DEvens

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    I think there must be something missing from your problem statement. What is it you are supposed to calculate? Also, I don't see an estimate of the shuttle's acceleration at the time the foam fell off. As a ball-park value it is probably something around 5g.

    Anyway, what is the initial velocity of the foam when it first loses contact with the shuttle? That gives you some idea of the force on the foam. Then, given the force, how do you get the acceleration at that moment?

    You then have a system of forces on a mass. Those forces depend on the velocity. You have a differential equation to solve.
     
  4. Aug 25, 2015 #3
    Does the drag force make the foam go faster than the shuttle or slower?
    If you change the word accelerating to decceleration, or negative acceleration.....

    An object can accelerate in the same direction as its velocity increasing its speed, or in the opposite direction as its velocity which should do what?
     
  5. Aug 25, 2015 #4

    HCD

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    The problem is to derive the equations that describe the velocity as well as the distance travelled by the foam as functions of time, but my problem is to model the forces acting on the foam due to the air flow.

    I was thinking of two possibilities, one where I should model the force due to air flow using the data in the problem statement, including maybe the velocity of the shuttle, and another where I should model the trajectory of the foam as a straight line (as I feel like that is kind of indicated in the problem statement) and that should somehow get me to the force due to air flow.

    In either case the foam starts from rest in the inertial reference frame of the shuttle (I'm assuming it's inertial since only the velocity of the shuttle is given in the time interval under consideration), but in the latter case there is no acceleration in the direction normal to the trajectory meaning that in this direction the force due to air flow must cancel the force of gravity. But still that says nothing about the force due to air flow in the direction tangent to the trajectory. Does it?

    In the model I had in mind the only effect of the shuttle speeding past the foam is to generate air flow in which the foam accelerates. Are you suggesting otherwise?
     
  6. Aug 25, 2015 #5
    I have no idea what you mean by that statement.

    According to the problem, the velocity of the shuttle is given as Vshuttle = 2500 ft/sec. We can assume that is in the direction upwards away from the earth and not downwards. That may seem obvious but it is the first step even if one did not have to ponder too much about it, but it is important.

    If the shuttle is travelling upwards, in what direction is the air moving relative to the shuttle?
    If the shuttle is travelling downwards, in what direction is the air moving relative to the shuttle?
    Are they in the same direction in both cases?
    Hopefully you answer No.

    For either case is what direction is the drag on the foam? Is it always in the opposite direction to the mg force,?

    That is what you should be looking at. Consider a piece of foam that comes off when the shuttle is travelling upwards in the vacuum of space where there is no air.
    Does the y-location of the foam change wrt the shuttle?

    So, in air what force causes the fom to move relative to the shuttle?

    Working out the movement of the foam wrt an inertial shuttle due to the brief time period is an OK stategy. It would be the nearly the same thing as putting the shuttle in a vertical wind tunnel and blowing the air past the shuttle at 2500 ft/sec. In that case, it is easy to conceive of the foam accelerating from zero velocity wrt the shuttle to a value when it hits the wing.
     
  7. Aug 25, 2015 #6

    HCD

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    Are you saying that drag is what accelerates the foam relative to the shuttle in the first place? That means that in the first instant after the foam comes loose from the shuttle the forces acting on the foam are
    $$ \boldsymbol{F} = \left[ mg + \frac{1}{2} \rho (2500\textrm { ft/s})^2 C_D A \right] \boldsymbol{e}_j, $$
    right? But how do I model the forces for the remainder of time under consideration? And I'm still confused since the trajectory of the foam doesn't seem to be parallel to the trajectory of the shuttle, or why would the wing span be given in the problem statement?
     
  8. Aug 25, 2015 #7
    Initially, I would just assume that the foam moves straight back and hits the wing and treat the problem as 1-D as a first approximation.

    Since the shuttle is shaped somewhat as a triangle in one plane, the air has to move sideways as well as towards the rear along the surface of the shuttle, so it has an x and a y component and a resultant with somewhat greater velocity. A second approximation would determine the resultant air speed assuming a traingle given by the dimensions of the shuttle and assume the foam will be induced to follow the same path. Except that we are not given the dimensions of the shuttle, so from the picture one could use the 78 ft and from the top view, estimate the detach point and contact point by the transfer of the arrow, if the arrow head is the contact point. Length of arrow is 56 ft, and one should be able to work out the other dimensions of the foam motion, taking into account also that it appears to follow a path towards the top of the shuttle. That is the only reason I can see that they gave you the wingspan.
    Compare that to your first estimate of the foam velocity wrt the shuttle at contact.

    ( What is that ej thing mean, especially e )
    That should make sense, yes. From an observer, on the ground, at detachment the foam is moving upwars at the same speed as the shuttle. Gravity and drag should both be acting downwards, slowing the foam down in its ascent. From an observer on the shuttle, gravity and drag are both accelerating the foam downwards. Make sense.

    One way to "solve" the problem is to assume a constant drag force ( from the 2500 ft/sec velocity ) just to see where you are at with the minimum time to impact and maximum velocity at impact.

    And then go on to the differential equation , which DEvans suggested. He should be able to help you with setting up.
     
  9. Aug 26, 2015 #8

    HCD

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    Okay, let's simplify the problem to 1-dimension and assume that drag due to air flow is constant. Then in the first instant after the foam comes loose from the shuttle the force ## F_0 ## acting on the foam can be modelled as
    $$ F_0 = mg + \frac{1}{2} \rho (2500 \textrm{ ft/s})^2 C_D A. $$
    But as the foam starts to speed towards the earth it's velocity relative to the shuttle ## v ## starts to be non-negligible and as a consequence drag starts to act also in the direction away from the earth, so for the time under consideration we could model the forces acting on the foam as
    $$ F = mg + \frac{1}{2} \rho (2500 \textrm{ ft/s})^2 C_D A - \frac{1}{2} \rho v^2 C_D A. $$
    Terminal velocity ## v_f ## is characterised by the following equation,
    $$ 0 = g + \frac{1}{2} \rho (2500 \textrm{ ft/s})^2 \frac{C_D A}{m} - \frac{1}{2} \rho v_f^2 \frac{C_D A}{m}. $$
    Which can be solved to get the terminal velocity as follows,
    $$ v_f = \sqrt{\frac{2(mg + F_a)}{C_D A \rho}} = \sqrt{\frac{2 \cdot 21554\textrm{ lb ft/s}^2}{C_D A \rho}} = 1770\textrm{ ft/s}, $$
    where ## F_a ## is the force due to air flow. This allows us to express the acceleration as a function of velocity introducing the terminal velocity as
    $$ a(v) = 10777 \textrm{ ft/s}^2 \left[1 - \left( \frac{v}{v_f} \right)^2 \right], $$
    where ## 10777 \textrm{ ft/s}^2 ## is ## (mg + F_a)/m ##.

    Now, using the relationship
    $$ t(v) = t_0 + \int_{v_0}^v \frac{dv}{a(v)} $$
    we can write
    $$ t(v) = \frac{1}{10777 \textrm{ ft/s}^2} \cdot \int_{0}^v \frac{dv}{1 - (v/v_f)^2} = \frac{v_f}{2 \cdot 10777 \textrm{ ft/s}^2} \cdot \ln \left( \frac{v_f + v}{v_f - v}\right), $$
    which can be solved for ## v ## to obtain
    $$ v(t) = v_f \cdot \frac{e^{2 \cdot \left( 10777 \textrm{ ft/s}^2 \right) \cdot t / v_f} - 1}{e^{-2 \cdot \left( 10777 \textrm{ ft/s}^2 \right) \cdot t / v_f} + 1}. $$
    We could even obtain the velocity as a function of the travelled distance from the relationship
    $$ s(v) = s_0 + \int_{v_0}^v \frac{v \cdot dv}{a(v)} $$
    by writing
    $$ s(v) = \frac{1}{10777 \textrm{ ft/s}^2} \int_0^v \frac{v \cdot dv}{1 - (v/v_f)^2} = - \frac{v_f^2}{2 \cdot 10777 \textrm{ ft/s}^2} \cdot \ln \left( 1 - (v/v_f)^2 \right). $$
    Which can be solved for ## v ## to obtain
    $$ v(s) = v_f \sqrt{1 - e^{2 \cdot \left( 10777 \textrm{ ft/s}^2 \right) \cdot s / v_f^2}}. $$
    Finally we can evaluate this last function for the flight path length of the foam in the air stream,
    $$ v(56 \textrm{ ft}) = 1001 \textrm{ ft/s}. $$
    But the answer given to the velocity of the foam when it strikes the wing is about 840 fps, which means that something is wrong with this model.
     
    Last edited: Aug 26, 2015
  10. Aug 26, 2015 #9

    billy_joule

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    No that's not correct.
    There a single drag force due to the relative velocity of the foam in the fluid. It is in the opposite direction to the fluid velocity relative to the foam.
    That is, the drag force is downward, towards earth.

    To consider terminal velocity as you did, when the foam block first detaches and is travelling upward makes no sense, there is no terminal velocity in this direction. No steady state can be reached, all forces are acting downward.
    I haven't done the math but I would bet the farm the foam is still travelling upward when it hits the shuttle.
    If the foam did not hit the shuttle (or bounced off) it would lose it's upwards (away from earth) velocity due to the downward drag and gravity forces and come to stop.
    It would then start to descend towards the earth, only then will an upward drag force exist and only then will terminal velocity become relevant.
     
  11. Aug 26, 2015 #10

    HCD

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    That all makes perfect sense, thanks. But then how would you model the problem?
     
  12. Aug 26, 2015 #11

    billy_joule

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    I don't know what the problem is, there's no question in the problem statement!
    I assume you need to find relative impact velocity, which is the change in velocity of the foam during the 0.15 seconds of flight time during which gravity and drag act on it.
    Always start with a free body diagram, it would've saved a lot of time in this case. Then develop the differential equation.

    If you aren't familiar with Diff. eqns. you could make some approximations. You know the distance the foam travelled relative to the shuttle and the shuttle speed so can find the distance travelled by the foam relative to the air/earth, along with the 0.15 s travel time you can find the foams average velocity. You can make some assumption about the deceleration of the foam and get an impact velocity. An iterative solution in excel should be pretty straight forward.
    Of course, this method can be done without the use of the drag equation which misses the point of the exercise..
    ..
     
  13. Aug 31, 2015 #12

    HCD

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    Here is what I understood so far.

    In the first instant under consideration the foam has zero velocity relative to the shuttle and a velocity of 2500 ft/s relative to the atmosphere. Still in that instant, the foam is experiencing a force due to gravity as well as drag caused by the relative motion of 2500 ft/s between the foam and the atmosphere. I could model the forces in that first instant as
    $$ F_0 = mg + \frac{1}{2} \rho (2500 \textrm{ ft/s})^2 C_D A, $$
    where ## \rho ## is the density of the atmosphere, ## C_D ## is the cross sectional area of the foam and ## A ## is the drag coefficient.

    In the next instant the force due to gravity acts the same but the drag force on the foam changes since the velocity of the foam relative to the atmosphere changes. My question is how do I model the forces throughout the time under consideration? Not just in the first instant. Could I model the forces as
    $$ F_0 = mg + \frac{1}{2} \rho (2500 \text{ ft/s} - v)^2 C_D A? $$
    where ## v ## is the velocity of the foam relative to the shuttle?
     
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