PSI vs MPH Calculation: Force for Wind Resistance

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The discussion centers on calculating the force needed to counteract wind resistance at 80 miles per hour, specifically focusing on the role of compressed air. Participants clarify that the inquiry involves determining the necessary PSI to combat wind over a distance of 3-4 meters. The application of compressed air is debated, with suggestions including its use in air motors or sails. It is recommended to utilize Bernoulli's equation for calculating velocity pressure to find the required force. Overall, the conversation highlights the complexities of applying compressed air to counteract wind resistance effectively.
norm88j
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Hello PF,

I would like to know what is the calculation for determing the amount of force it would take to combat PSI of the wind in a vehicle moving at 80 mile per hour.
 
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Are you talking about how to calculate wind resistance for an automobile? Is that the question?
 
yes, more importantly compressed air going in the opposite direction. How much compression or compressed psi would be necessary tocombat the wind 3-4 meters?
 
Where do you envision applying compressed aire to oppose the wind? How much will be required seems like it might depend upon how you will use it. Will you use it in an air motor working on the wheels to power the wheels? Will you apply it to a sail? Just what are you thinking? What do you mean when you add "3-4 meters"? This really does not make much sense.
 
the opposing air would come from a motorized compressed apparatus or hose.
 
Use the simple form of bernoulli's equation for velovity pressure.
 
I have Mass A being pulled vertically. I have Mass B on an incline that is pulling Mass A. There is a 2:1 pulley between them. The math I'm using is: FA = MA / 2 = ? t-force MB * SIN(of the incline degree) = ? If MB is greater then FA, it pulls FA up as MB moves down the incline. BUT... If I reverse the 2:1 pulley. Then the math changes to... FA = MA * 2 = ? t-force MB * SIN(of the incline degree) = ? If FA is greater then MB, it pulls MB up the incline as FA moves down. It's confusing...
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