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# Calculate objects position in water affected by buoyancy

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### #21Hairie  Members   -  Reputation: 109

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Posted 14 May 2012 - 04:46 PM

Okas, I get that...but how is it, if the bouyancy still is less then the gravity force on the object? How does that affect the resulting acceleration? Because it already slows down, before its buoyancy is bigger than its gravity force ...or not?

### #22Bacterius  Crossbones+   -  Reputation: 9054

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Posted 14 May 2012 - 04:50 PM

The acceleration can never decrease beyond g (=-9.8 m/s) since the buoyancy force is *always* upwards however the velocity will of course. Does it make sense?

The slowsort algorithm is a perfect illustration of the multiply and surrender paradigm, which is perhaps the single most important paradigm in the development of reluctant algorithms. The basic multiply and surrender strategy consists in replacing the problem at hand by two or more subproblems, each slightly simpler than the original, and continue multiplying subproblems and subsubproblems recursively in this fashion as long as possible. At some point the subproblems will all become so simple that their solution can no longer be postponed, and we will have to surrender. Experience shows that, in most cases, by the time this point is reached the total work will be substantially higher than what could have been wasted by a more direct approach.

- Pessimal Algorithms and Simplexity Analysis

### #23Hairie  Members   -  Reputation: 109

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Posted 14 May 2012 - 04:58 PM

Yea it makes sense, but you said that if the bouyancy is bigger than its gravityforce only then it will begin to deaccelerate, but what is, if bouyancy is smaller than its gravityforce? Then it should also deaccelerate or not?

### #24jefferytitan  Crossbones+   -  Reputation: 2219

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Posted 14 May 2012 - 05:12 PM

If

buoyancy is smaller, it will decelerate still. Think about a ship sinking. It sinks slower than it would fall through air, even though it still sinks.

My advice is just to implement it and see how it goes. Even for light objects

buoyancy will be smaller than the gravity force *for a while*, e.g. until enough of it is submerged to provide a strong buoyancy force.

### #25Hairie  Members   -  Reputation: 109

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Posted 14 May 2012 - 11:23 PM

Okay, but, if buyancy is bigger than its gravityforce it should actually start rising with the resulting accelreation and not decelrate by (g-resulting acceleration)? Because if so, it would decelerate even if it had its maximuim bouyancy force. For example if a body had the density 0.75 and a volume 0f 1000, its gravityforce would be 7357.5N.
Its maximum buoyancy would be 9819. The resulting force would be 2452.5 N and the resulting max acceleration would be 3.27 m/s², so it should rise by that acceleration and not sink by(g-resulting acceleration), right?

Edited by Hairie, 14 May 2012 - 11:33 PM.

### #26Bacterius  Crossbones+   -  Reputation: 9054

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Posted 14 May 2012 - 11:59 PM

I think you are confusing velocity and acceleration (if you are not, apologies). Acceleration is (in the least technical sense of the term) the tendency of an object to go in a particular direction. Velocity is the direction the object is actually going. Acceleration affects velocity which affects position (but acceleration never directly affects position).

Okay, but, if buyancy is bigger than its gravityforce it should actually start rising with the resulting accelreation and not decelrate by (g-resulting acceleration)? Because if so, it would decelerate even if it had its maximuim bouyancy force. For example if a body had the density 0.75 and a volume 0f 1000, its gravityforce would be 7357.5N.
Its maximum buoyancy would be 9819. The resulting force would be 2452.5 N and the resulting max acceleration would be 3.27 m/s², so it should rise by that acceleration and not sink by(g-resulting acceleration), right?

It won't rise automatically. It will slowly lose downwards velocity, and then start rising back up when the velocity flips signs under the influence of the upwards acceleration.

Imagine a car going at constant speed on a road. Its acceleration is zero. When the car is braking, the brakes force the road to apply friction to the car, which creates a force pointing in the opposite direction of the car's motion. When the car starts losing speed, its acceleration is in fact negative. At some point, its velocity becomes zero, and the car stops. At this point, the car should start going backwards under the negative acceleration it is undergoing (in real life it doesn't happen because of how car brakes work although you can glimpse it by seeing how cars are kind of pushed backwards a bit after quickly decelerating, but are stopped by static friction from the road and brakes).

Short story long, object's don't move from sheer acceleration - acceleration influences velocity which then causes the object to move.

Edited by Bacterius, 15 May 2012 - 04:25 AM.

The slowsort algorithm is a perfect illustration of the multiply and surrender paradigm, which is perhaps the single most important paradigm in the development of reluctant algorithms. The basic multiply and surrender strategy consists in replacing the problem at hand by two or more subproblems, each slightly simpler than the original, and continue multiplying subproblems and subsubproblems recursively in this fashion as long as possible. At some point the subproblems will all become so simple that their solution can no longer be postponed, and we will have to surrender. Experience shows that, in most cases, by the time this point is reached the total work will be substantially higher than what could have been wasted by a more direct approach.

- Pessimal Algorithms and Simplexity Analysis

### #27Hairie  Members   -  Reputation: 109

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Posted 15 May 2012 - 05:52 AM

yes your right, even if the resulting acceleration is > 0, the velocity will still be negative and only shortly after velocity will become positive and the object will rise again

Edited by Hairie, 15 May 2012 - 05:53 AM.

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