Does a Glider Complete Its Loop Starting at 60 m/s with 30 m/s² Acceleration?

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The discussion revolves around calculating the motion of a glider completing a loop while flying horizontally at 60 m/s with an initial centripetal acceleration of 30 m/s². Participants emphasize the importance of understanding Newton's laws and the relationship between lift force, speed, and centripetal acceleration. The conversation highlights the need for numerical methods to solve the differential equations governing the glider's motion, particularly under the influence of gravity. Suggestions include breaking down the time into small increments to track changes in position, velocity, and acceleration. Ultimately, the consensus is that the glider can complete the loop, returning to its starting position, provided the calculations account for the forces involved correctly.
  • #31
Height in your case is y, not x. Good otherwise.
 
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  • #32
y of course :)

Thank you, I'll check that out
 
  • #33
At the bottom of the loop where y = 0 and v = 60 m/s the KE is (1/2)mv^2 = 1800m J. At the top of the loop I have v=19.36m/s and y= 166.061, for a total energy of (1/2)mv^2 +mgh = 1814.8m J. I think the error is due to the way that the Euler method assumes for each incremental step that acceleration is constant in both x and y directions - it ignores the fact that the plane's angle has changed which would have the effect of slowing its forward velocity.
Should I include the angle somehow with this simplified equations?
 
  • #34
You have the error because Euler's method approximates a continuous function with a piecewise linear curve. This should have been explained to you in your "numerical methods" course. You cannot eliminate this error, but you can make it smaller by reducing your time step. Another source of error is the finite precision of computer arithmetic. This you cannot reduce at all (without taking some extreme measures, that go far beyond your little assignment).

Keeping track of total energy let's you estimate the error; your error is about 1%, which is not too bad, especially if you recall that you ignore drag completely, which will have a much greater effect on energy and other variables.
 
  • #35
I understand. Thank you Voko!

I'll now start finishing this program and start writing report.
I might have a questions or two for the process of evolving those equations.
I'll post the finished assignment/program here. Thanks to everyone who helped and participate in discussion.
 
  • #36
Somebody suggested to me that

a_x = ( 30 (centripetal acceleration) * v * v_x1 ) / 60*60
a_y = (( 30 (centripetal acceleration) * v * v_y1 ) / 60*60 ) - 9,8

would be correct.

And
a_x = -39.8 * V_t * Vy_1 / (60 * 60);
a_y = 39.8 * V_t * Vx_1 / (60 * 60) - 9.8;
is wrong.

Why should it be like that?
Sorry guys, for dummy question, I really respect your patience with me, but .. How can I learn to understand these final equation? Where to start? Where do they come from exactly?
Voko your explanation for upper difficult equations makes sense for a start.
Where do I begin to study those final simplified equations?

For example - POSITION:
I see why there is '' x_2 = x_1 + (Vx_1) * delta_t ''
V * t = x
But how do I know V_x is constant?

VELOCITY:
Vx_2 = Vx_1 + Ax * delta_t
a * t = V ... I agree, but this time we are using equation for accelerated motion. Why? If we used an equation for constant velocity for position?

ACCELERATION:
Ax = -39.8 * V_t * Vy_1 / (60 * 60) or a_x = ( 30 (centripetal acceleration) * v * v_x1 ) / 60*60
This one I don't understand at all.
 
Last edited:
  • #37
StudentTM said:
Somebody suggested to me that

a_x = ( 30 (centripetal acceleration) * v * v_x1 ) / 60*60
a_y = (( 30 (centripetal acceleration) * v * v_y1 ) / 60*60 ) - 9,8

would be correct.

As you explained earlier yourself, the initial acceleration is 30 m/s2, and because we have -9.8 m/s2 due to gravity, the lift coefficient must be 39.8 / (60*60). So these equations cannot be correct.

a_x = -39.8 * V_t * Vy_1 / (60 * 60);
a_y = 39.8 * V_t * Vx_1 / (60 * 60) - 9.8;
is wrong.

The coefficient is correct. The signs are also correct, which should be intuitively clear: as the pilot pulls up, the horizontal velocity must start decreasing, while the vertical velocity must go up.

Sorry guys, for dummy question, I really respect your patience with me, but .. How can I learn to understand these final equation? Where to start?

Newton's second law, gravity, lift. High-school material.

But how do I know V_x is constant?

It is not constant. But when you integrate differential equations using Euler's method, you assume that during the very small time interval of each integration step, all your variables do not change appreciably. This is the source of the error intrinsic to the method. You should have been paying attention to your professor, this has certainly been explained in the class.
 
  • #38
voko said:
It is not constant. But when you integrate differential equations using Euler's method, you assume that during the very small time interval of each integration step, all your variables do not change appreciably. This is the source of the error intrinsic to the method. You should have been paying attention to your professor, this has certainly been explained in the class.

Ohh, that explains a lot :) Thanks .. Yes I should be paying more attention at class for sure ;)
 

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