Model Rocket- What does the 'v' in drag=kv^2 really mean?

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Discussion Overview

The discussion revolves around the interpretation of the variable 'v' in the drag equation \( \text{drag} = kv^2 \) as it applies to predicting the peak altitude of a model rocket. Participants explore the implications of different definitions of velocity in the context of forces acting on the rocket, including thrust, gravity, and drag.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions whether 'v' should represent the velocity of the rocket considering the effects of gravity and air resistance or just the velocity from the motor without those effects.
  • Another participant clarifies that velocity is defined as the rate of change of position over time and emphasizes that the motor provides a force, not a direct velocity.
  • A participant notes that 'v' should be the velocity with respect to the surrounding air, suggesting that drag is dependent on this relative velocity.
  • A later reply indicates that the participant has calculated velocity based on ground testing and considers the challenge of incorporating drag due to having two unknowns: the force due to drag and the actual velocity.
  • There is a reiteration that in this context, "drag" and "air resistance" are synonymous.

Areas of Agreement / Disagreement

Participants express differing views on the appropriate definition of 'v' in the drag equation, indicating that there is no consensus on this aspect of the discussion.

Contextual Notes

Participants mention the complexity of incorporating drag into calculations due to the presence of multiple unknowns, which may affect the accuracy of predictions regarding peak altitude.

Andy24
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Hello! I am predicting the peak altitude of a model rocket based on some ground tests. I know that Total Force on the rocket=Thrust-mg-kv^2 but am stuck as to which value for 'v' to use? Is it the velocity of the rocket with the effects of gravity and air resistance taken into account or the velocity purely from the motor (without any air resistance or gravity taken into account)?

Your help is appreciated, thanks.
 
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The velocity at any given time is the rate of change of position over time, how fast the object is actually going. There is no such thing as "the velocity purely from the motor": the motor does not give a velocity, it gives a force. The velocity will be the result of all the forces acting on the object (and its previous velocity if it is already moving).
 
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DrClaude said:
That's reminds me that I forgot to add that it is the velocity with respect to the surrounding air.

Thanks for your response.. I understand now. I actually calculated the velocity based on some ground testing (where I found out the impulse of the motor) and was hoping to incorporate drag.. This may give you context: https://www.physicsforums.com/threads/coefficient-of-drag-on-a-model-rocket.871810/
Thanks for your help though, I think I might just leave incorporating drag into my calculations as I have 2 unknowns- the force due to drag as well as the actual velocity. Thanks again,
Andrea
 
Andy24 said:
Hello! I am predicting the peak altitude of a model rocket based on some ground tests. I know that Total Force on the rocket=Thrust-mg-kv^2 but am stuck as to which value for 'v' to use? Is it the velocity of the rocket with the effects of gravity and air resistance taken into account or the velocity purely from the motor (without any air resistance or gravity taken into account)?
In this context, the terms "drag" and "air resistance" are synonymous.
 

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