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Integrate Acceleration 

Rick Sellens
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This video is about Integrating measured Acceleration data to get velocity and position.

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8 сен 2024

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Комментарии : 16   
@ryannorooz
@ryannorooz 4 года назад
you just saved me a lot of headache...! i can't thank you enough❤❤❤
@rybread5718
@rybread5718 3 года назад
Thanks man! Great video!
@krishnavamshi9340
@krishnavamshi9340 5 месяцев назад
If im calculating Displacement along z axis should i consider anything additionally as gravity is acting along that axis
@robv3872
@robv3872 2 года назад
What types or approaches could one read about for removing error when using the IMU approach? Drift and computational errors...is there a standard approach used for this?
@OctaRudin
@OctaRudin 8 месяцев назад
I tried this concept in arduino nano. but, the velocity/position gets bigger and bigger. what should I do? thanks.
@akshaygund9068
@akshaygund9068 6 лет назад
tried this but very inaccurate result due to small errors in acclerometer data .......there is no acclerometer which gives accurate accleration.
@RickSellens
@RickSellens 5 лет назад
That's sort of the point. Inertial measurements will be subject to drift and will always need some other source to bring them back from those errors that magnify over time.
@themanthis837
@themanthis837 8 лет назад
where is the programming example`?
@RickSellens
@RickSellens 8 лет назад
In the course notes for registered students. Sorry
@Say252
@Say252 5 лет назад
V_2 = V_1 +( A2 - A1)/dT not A1 + A2 !
@RickSellens
@RickSellens 5 лет назад
You are mistaken. (A1+A2)/2 is the average acceleration x dT will give change in velocity.
@Say252
@Say252 5 лет назад
@@RickSellens ohh, sorry! My fault
@henryh.8381
@henryh.8381 5 лет назад
@@RickSellens Hello Rick, i Don't understand why choising the average from derive. In case of a big deceleration for exemple, at t(5) we could have Acc = +2.5m/s² and at t(6) = -2.5m/s² -> (2.5+ (-2.5))/2 = 0 in Derive like Felipe, ((-2.5)-2.5) / dt 0
@RickSellens
@RickSellens 5 лет назад
Henry H. If A is positive for the first half of the time step, then negative for the second half of the time step, the net change in velocity should be zero. The shaded trapezoids are the areas being calculated. Between t3 and t4 the triangles will tend to cancel out.
@henryh.8381
@henryh.8381 5 лет назад
@@RickSellens Thanks for your reply. now i need time to understand it (because my English stayed to low level). Maybe, i'll be back, lol :)
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