Looking for an intelligent argument. (Terminal Velocity)

I've obviously misinterpreted barnets post. If we're travelling in a plane at a constant speed at a constant height above Sea Level, what is the reading on the acclerometer? And what is it measuring?

I've obviously misinterpreted barnets post. If we're travelling in a plane at a constant speed at a constant height above Sea Level, what is the reading on the acclerometer? And what is it measuring?

Quote from: Fletch on September 01, 2008, 05:39:39 AM

I've obviously misinterpreted barnets post. If we're travelling in a plane at a constant speed at a constant height above Sea Level, what is the reading on the acclerometer? And what is it measuring?

It's OK, it gets a bit confusing if you start including calibrating it to read zero on the ground and so on.  I am assuming that on the ground/in the plane (constant altitude), the accelerometer is measuring the acceleration due to the contact force of you standing on the floor. What is happening, essentially, is that a test mass is 'falling' towards the ground on one end of a spring. The other end of the spring is attached to you, which can not fall since you are standing on the ground, therefore it measures an extension of the spring and outputs an acceleration.

Now imagine you have just jumped out of the plane. Now the weight on the spring is falling towards the ground, but so are you (at exactly the same rate).  The spring does not extend and so no acceleration is measured. This will be the same in orbit or anywhere else where you are in free fall.  Once you are out of the plane, the drag from the air will start to increase, which means the accelerometer can start to extend again (giving a reading) until you reach terminal velocity and the reading is the same as on the ground.

Does this make sense?

Yes. Barnets post said that it measures acceleration in relation to itself, so I thought that meant it registered changes in acceleration and when travelling at a constant velocity it registered 0.

Either way, FE and RE would yield the same reading.

Yeah, that was my point - barnet's only real mistake was assuming that an accelerometer could measure local gravitational field strength (local g) while in free fall. Pretty much everything else stemmed from that. Hopefully that's resolved now (we'll have to wait for his next post to see).

Quote from: Matrix on September 01, 2008, 05:34:00 AM

The accelerometer will read as I have described in my previous post. It will change from reading the acceleration due to gravitation/UA in the plane (flying at constant altitude) to reading zero when you leave the plane. It then registers a gradual increase until, at terminal velocity, it measures the same as when you were on the ground/in the plane.

The key is to remember that the accelerometer cannot 'see' the ground, it only has internal workings to go on.  This explains why it can't sense gravitation in free-fall, only the effects of drag (which, at terminal velocity, are the same as the contact acceleration on the ground).

I've obviously misinterpreted barnets post. If we're travelling in a plane at a constant speed at a constant height above Sea Level, what is the reading on the acclerometer? And what is it measuring?

No accelerometer in a plane that I've have ever seen is measured in m/s^2.  What it does say is 1g. Because as you stand in the plane or sit your butt in the seat it feels no different than if your sitting your butt or standing on the ground.

But that is not what the gauge reads.  Which is what he asked.

I know they equal each other.  But when we go into 2g and 3g turns you have to be able to measure the Gs for structural limits.  Which are listed in Gs not m/s^2

Just admit im right.

Does that mean win for FE?

our acft dont have an m/s^2 gauge either...

our acft dont have an m/s^2 gauge either...

Quote from: RETftw on September 01, 2008, 08:58:24 AM

our acft dont have an m/s^2 gauge either...

What do you fly?

Ft Campbell?

Is there no limit to the amount of fail cbarnett is capable of?

im sorry - there was a question?

Or that you fail to grasp simple physics.

Quote from: TheEngineer on August 31, 2008, 10:00:12 PM

Quote from: cbarnett97 on August 31, 2008, 09:36:56 PM

that is the mistake theengineer is making and everyone is following his lead, you can not relate it to the earth until you have reached your point of equalibrium.

It's equilibrium, kid. 

Quote

Quote from: cbarnett97 on August 31, 2008, 03:20:53 PM

the whole idea of doing a free body diagram is to look at the forces acting on the body.

And like I said already, there is no balancing force.  Therefore there is a constant acceleration. 

I believe this is too complicated for you to follow.  I thought my derivation was simple enough for you.  Apparently, I gave you too much credit.  We are only interested in the velocity at a certain acceleration.  Is that simple enough for you?

so using the FE model what would an accelerometer show at t=2

A little under 9.8m/s^2.

I'm still waiting for you to show me where my derivation is wrong.