Center of gravity of objects on a flat earth
« on: August 19, 2020, 07:09:02 PM »
I’ve been reading about universal acceleration in the wiki and it’s a good alternative explanation for gravity for objects that are free falling.  But not everything that falls is in free fall.  Things that are already on the ground or supported in some way topple over, or fall, or drop all the time. According to conventional physics, it is because an objects center of gravity shifts or isn’t low enough to keep it upright.  Something with a low center of gravity is much more stable than something with a higher center of gravity and is less likely to fall.

Is there an FE explanation for why somethings are more stable than others since the conventional “center of gravity” concept wouldn’t apply (I’m assuming)?  What would cause something to be balanced or unbalanced on the flat earth?

Offline iamcpc

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Re: Center of gravity of objects on a flat earth
« Reply #1 on: August 19, 2020, 08:36:09 PM »
Is there an FE explanation for why somethings are more stable than others since the conventional “center of gravity” concept wouldn’t apply (I’m assuming)?  What would cause something to be balanced or unbalanced on the flat earth?


In both the RE and the FE models someone who is accelerating upward would be unable to tell the difference from this upward acceleration and being inside of a gravitational field. Here's a video explaining this.




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Offline JSS

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Re: Center of gravity of objects on a flat earth
« Reply #2 on: August 19, 2020, 09:36:49 PM »
Is there an FE explanation for why somethings are more stable than others since the conventional “center of gravity” concept wouldn’t apply (I’m assuming)?  What would cause something to be balanced or unbalanced on the flat earth?
In both the RE and the FE models someone who is accelerating upward would be unable to tell the difference from this upward acceleration and being inside of a gravitational field. Here's a video explaining this.



Thats a good video, it explains things quite clearly. But it does leave out one important point, the thought experiment in the Equivalence Principle has some conditions. You must be in a small space, and you can't look outside the elevator.

This is because there are differences between a large gravity well and an accelerating spaceship. The most obvious one would be that gravity weakens the further from the floor you get in a gravity well, but does not change in a spaceship.

Back in Einsteins day this was considered too small to measure, but today we have gravimeters that can tell the difference even in an elevator so that thought experiment is a bit out of date.

But the Equivalence Principle still holds, acceleration and gravity are the same thing. It's just now we are MUCH better at being able to tell what's creating that force.

Under Universal Acceleration, all the laws of gravity we currently expect to see here on Earth would still apply so the answer to all the balance and center of gravity questions would be the same and not an issue until you start looking VERY closely or use large scale experiments. Then you would start seeing differences.

https://en.wikipedia.org/wiki/Equivalence_principle

"An observer in a windowless room cannot distinguish between being on the surface of the Earth, and being in a spaceship in deep space accelerating at 1g. This is not strictly true, because massive bodies give rise to tidal effects (caused by variations in the strength and direction of the gravitational field) which are absent from an accelerating spaceship in deep space. The room, therefore, should be small enough that tidal effects can be neglected."

Re: Center of gravity of objects on a flat earth
« Reply #3 on: August 19, 2020, 11:04:45 PM »
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In both the RE and the FE models someone who is accelerating upward would be unable to tell the difference from this upward acceleration and being inside of a gravitational field. Here's a video explaining this.



I understand the equivalence principle and see how it is a good alternate theory of gravity for objects in free fall.  But I don't think it applies to my question.

The ground can't move up to meet something that's already on the ground when it falls.  If you're already standing on the ground and lean over far enough, you lose your balance and fall. Technically, you rotate. The ground doesn't move up to meet you if your feet never leave the ground.  IF your feet are still in contact with the ground, your whole body should continue to move up with the ground.

And even if your feet do leave the ground, the equivalence principle still doesn't explain why lose your balance in the first place.  The mainstream explanation is that your center of gravity is no longer directly over your support and gravity will act on the part of your body that has the most mass.  My assumption is that flat earth wouldn't recognize an object's center of gravity as being a real thing.

Think of a simple balance scale that is resting on the ground.  The ground is accelerating up, taking the scale with it.  Put a heavy rock on one side and it will "fall".  It might not touch the ground but it still falls, all the while the scale as a whole is also moving up with the ground.

Or a teeter totter.  If a heavier kid is on one side, that side will fall, maybe touching the ground or not, but the whole teeter totter is rising at the same rate, so why would one side fall and not the other?

Offline iamcpc

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Re: Center of gravity of objects on a flat earth
« Reply #4 on: August 20, 2020, 03:16:48 PM »
I understand the equivalence principle and see how it is a good alternate theory of gravity for objects in free fall.  But I don't think it applies to my question.

I think it does.

The ground can't move up to meet something that's already on the ground when it falls.

Yes it can

  If you're already standing on the ground and lean over far enough, you lose your balance and fall. Technically, you rotate. The ground doesn't move up to meet you if your feet never leave the ground.


If you were in a closed system, like on a planet, you would be unable to tell if your face was slamming into the ground or if the ground was slamming into your face when you fall on your face.



And even if your feet do leave the ground, the equivalence principle still doesn't explain why lose your balance in the first place.  The mainstream explanation is that your center of gravity is no longer directly over your support and gravity will act on the part of your body that has the most mass.  My assumption is that flat earth wouldn't recognize an object's center of gravity as being a real thing.


The center of gravity is the average location of the weight of an object. Having defined that lets do this.


Losing your balance and tipping over is, as you defined it, when "your center of gravity is no longer directly over your support"


An object tips over when The average location of the weight of an object is no longer over the support.

In the gravity system An object tips over when The average location of the weight of an object is no longer over the support.
In an upward accelerating system An object tips over when The average location of the weight of an object is no longer over the support.


Or a teeter totter.  If a heavier kid is on one side, that side will fall, maybe touching the ground or not, but the whole teeter totter is rising at the same rate, so why would one side fall and not the other?

Force. The heavier kid on the teeter totter has more downward force therefore it moves down. In the acceleration system F = MA. Force = Mass * Acceleration.

The difference is that, in the gravity system, if a block is sitting on the ground not moving, it still have gravitational force even though it's not accelerating.

Re: Center of gravity of objects on a flat earth
« Reply #5 on: August 20, 2020, 04:24:35 PM »
Like IAMCPC said. 

Don't get hung up on the name "Centre (or Center) of Gravity", just because it has "gravity" in the title.  Engineers and scientists use "Center of Mass". 

It doesn't matter whether the acceleration is due to gravity, ball hit by a bat, bullet in a gun, or something on the floor of an aircraft pulling "g".  As long as the accelerating force is distributed around the Centre of Mass (C of M) it's stable. 

In the teeter totter (I'm assuming Brit English see-saw), lets say Skinny Kid is on the left, Fat Kid on the right.  (Can I say Fat kid?).  The pivot is in the middle, C of M is to the right-of-middle. 

If the planet is accelerating up, the force is applied to the pivot, which is left of the C of M. 

If Earth's Gravity is applying a down-force to the C of M, the reaction is felt upwards at the pivot, which is, again, left of the C of M. 

Either way, the 2 forces are not aligned so the outfit rotates clockwise until Fat Kid's butt hits the planet, at which time, C of M is now supported by the pivot and his butt; stability restored. 

Gravity? Acceleration?  Doesn't matter. 

Re: Center of gravity of objects on a flat earth
« Reply #6 on: August 20, 2020, 05:18:17 PM »
Hello. I have recently been presented with proof that the earth is flat. For all you who have done research on this topic, please guide me to a discussion board where I may find the information I need. I have a degree in science(I know, big mistake). I am not religious, but very interested in flat earth theory(or fact, whatever you call it). Thanks.

Re: Center of gravity of objects on a flat earth
« Reply #7 on: August 20, 2020, 07:24:49 PM »
Hello. I have recently been presented with proof that the earth is flat. For all you who have done research on this topic, please guide me to a discussion board where I may find the information I need. I have a degree in science(I know, big mistake). I am not religious, but very interested in flat earth theory(or fact, whatever you call it). Thanks.

I'm gonna do what everyone here is gonna do: link you to the wiki https://wiki.tfes.org/The_Flat_Earth_Wiki
there you can read proofs that sometimes contradict themselves and also are not accepted by everyone that thinks the earth is flat.

Even though I really would like to see the proof you've got there.

Re: Center of gravity of objects on a flat earth
« Reply #8 on: August 20, 2020, 11:37:23 PM »
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It doesn't matter whether the acceleration is due to gravity, ball hit by a bat, bullet in a gun, or something on the floor of an aircraft pulling "g".  As long as the accelerating force is distributed around the Centre of Mass (C of M) it's stable.


But what if it isn't distributed around the COM?  That's kind of my point.  The COM is where gravity will exert the most pull, because that is where the object's mass is concentrated.  But from what I can tell from the wiki, the "accelerator force" must be evenly distributed. It doesn't vary according to position.

If two kids are of equal weight on a teeter totter with the pivot in the middle, it will balance.  But if you move the pivot, you shift the COM and one side increases in mass.  Greater mass means greater gravitational pull.  Gravity is pulling harder on one side than on the other and causes the more massive side to fall. If the accelerator force is what causes it to fall, then the accelerator force would have to be greater on one side than on the other.

There's nothing in the wiki to suggest that greater mass means greater accelerator force and I don't even know how that would work.  The ground on one side would have to be accelerating at a different rate than the other.

Re: Center of gravity of objects on a flat earth
« Reply #9 on: August 21, 2020, 08:06:37 AM »
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It doesn't matter whether the acceleration is due to gravity, ball hit by a bat, bullet in a gun, or something on the floor of an aircraft pulling "g".  As long as the accelerating force is distributed around the Centre of Mass (C of M) it's stable.


But what if it isn't distributed around the COM?  That's kind of my point.  The COM is where gravity will exert the most pull, because that is where the object's mass is concentrated.  But from what I can tell from the wiki, the "accelerator force" must be evenly distributed. It doesn't vary according to position.

If two kids are of equal weight on a teeter totter with the pivot in the middle, it will balance.  But if you move the pivot, you shift the COM and one side increases in mass.  Greater mass means greater gravitational pull.  Gravity is pulling harder on one side than on the other and causes the more massive side to fall. If the accelerator force is what causes it to fall, then the accelerator force would have to be greater on one side than on the other.

There's nothing in the wiki to suggest that greater mass means greater accelerator force and I don't even know how that would work.  The ground on one side would have to be accelerating at a different rate than the other.


well actually when the teeter totter is  fixed on the ground of the accelerating box it will accelerate with the same force. So it would behave like gravity. For example if you put a plank on your car front center with a hinge ( in driving direction) and attach two equal weights when accelerating you  should see that the plank will not move to one side or the other (in reality this would be very difficult to do because there are many things interfering with your setup) but when you offset the center of the plank to one side and accelerate then it will move to the other side because of leverage and mass distribution.

Offline iamcpc

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Re: Center of gravity of objects on a flat earth
« Reply #10 on: August 21, 2020, 06:52:30 PM »
But what if it isn't distributed around the COM?  That's kind of my point.  The COM is where gravity will exert the most pull, because that is where the object's mass is concentrated.  But from what I can tell from the wiki, the "accelerator force" must be evenly distributed. It doesn't vary according to position.


if an object being pulled down by a  gravitational field is supported by something in which is COM is not over the support it topples over.

if an object being pulled down by acceleration is supported by something in which is COM is not over the support it topples over.

I don't know how much clear this can be.

If a heavy child is on one side of a teeter totter and a smaller child is on the other side of a teeter totter and they are both being pulled down by a gravitational field then the heavy child will be pulled down.
If a heavy child is on one side of a teeter totter and a smaller child is on the other side of a teeter totter and they are both being pulled down by acceleration then the heavy child will be pulled down.

Re: Center of gravity of objects on a flat earth
« Reply #11 on: August 21, 2020, 09:33:12 PM »
The OP needs to study Mechanics 101. He's ignoring turning effects when a force is applied away from the CoM (Centre of Mass). Gravity is considered as being applied at the CoM, not anywhere else. A resting body under gravity whose CoM is not over or between points of contact with the ground will topple and fall. A body in contact with an upwardly accelerating plane whose CoM is not over or between points of contact with the plane will also topple or fall.

The teeter totter example is easier understood if you think of Calvin (weight 55lbs) and Moe (weight 120lbs) sitting on the seats at either end, so each is 6ft away from the pivot. The result is exactly the same as placing a weight equal to the difference between Calvin and Moe (120-55=65lbs) on Moe's seat. What do you think would happen if you placed a 65lb weight on one seat of an empty teeter totter? Doesn't matter if the teeter totter is on a massive globe or an upwardly accelerating plane, the result is the same.

Flat Earthery (and general relativity's Equivalence Principle) has nothing to do with this, it's simple mechanics.
« Last Edit: August 21, 2020, 09:35:02 PM by Longtitube »

Re: Center of gravity of objects on a flat earth
« Reply #12 on: August 22, 2020, 05:11:11 AM »
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if an object being pulled down by acceleration is supported by something in which is COM is not over the support it topples over.

You can’t be pulled down by acceleration that is pushing up.

As was pointed out earlier in the thread…F=M*A, which is true but not the complete picture.
Force and acceleration are vectors and have both direction and magnitude.  Simply assigning a magnitude to a force doesn’t tell you squat about how the force is affecting something.  Is it moving it up? Is it moving it down? Sideways? Crushing it?

In order to answer those questions, you have to assign a direction to the acceleration. Assign upward (or positive) direction to the acceleration, as you would have on a flat earth, you end up with an upward force. Assign a downward (or negative direction), as you would have with gravity, you end up with a downward force.  How is an upward force pulling something down?

The bottom line is that when something is balanced, is simply means that gravity is acting equally on all parts of an object.  By that definition, I don’t see how anything could ever be unbalanced on a flat earth, considering that the acceleration force is constantly and consistently accelerating everything with the same amount of force.  Even if that weren’t the case, I don’t see how it could cause something to fall “down”, when the force itself is up. Any downward force that would come from a supported object would be the normal force, and would it would only be equal to the upward force.  No net force, means no movement in any direction. 



Offline iamcpc

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Re: Center of gravity of objects on a flat earth
« Reply #13 on: August 24, 2020, 04:02:23 PM »
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if an object being pulled down by acceleration is supported by something in which is COM is not over the support it topples over.

You can’t be pulled down by acceleration that is pushing up.

As was pointed out earlier in the thread…F=M*A, which is true but not the complete picture.
Force and acceleration are vectors and have both direction and magnitude.  Simply assigning a magnitude to a force doesn’t tell you squat about how the force is affecting something.  Is it moving it up? Is it moving it down? Sideways? Crushing it?




Are you seriously arguing semantics? Fine then. I'll correct my previous statement.


If a support is pushing up against an object because of gravity in which is COM is not over the support it topples over.

If a support is pushing up against an object because of acceleration in which is COM is not over the support it topples over.

I don't know how much clear this can be.

If a heavy child is on one side of a teeter totter and a smaller child is on the other side of a teeter totter and the teeter totter is pushing up against them because of a gravitational field then the heavy child will be pulled down.
If a heavy child is on one side of a teeter totter and a smaller child is on the other side of a teeter totter and the teeter totter is pushing up against them because of acceleration then the heavy child will be pulled down.



In order to answer those questions, you have to assign a direction to the acceleration. Assign upward (or positive) direction to the acceleration, as you would have on a flat earth, you end up with an upward force. Assign a downward (or negative direction), as you would have with gravity, you end up with a downward force.  How is an upward force pulling something down?

Because of the laws of inertia. When I hit the gas in my car i'm pulled back into the seat of my car, opposite the direction of acceleration.

By that definition, I don’t see how anything could ever be unbalanced on a flat earth, considering that the acceleration force is constantly and consistently accelerating everything with the same amount of force.

Acceleration <> force.

Force = Mass * acceleration. So when accelerating upwards at the same rate objects with more mass have more force.



Even if that weren’t the case, I don’t see how it could cause something to fall “down”, when the force itself is up. Any downward force that would come from a supported object would be the normal force, and would it would only be equal to the upward force.  No net force, means no movement in any direction.

Sounds to me like you should go talk to an engineering professor or physics professor to see if they can explain it better. Everyone here is in agreement about this.

Re: Center of gravity of objects on a flat earth
« Reply #14 on: August 26, 2020, 04:04:03 AM »
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Because of the laws of inertia. When I hit the gas in my car i'm pulled back into the seat of my car, opposite the direction of acceleration.

You will only be pulled back with a force that is equal the acceleration.  Inertial forces are opposite and equal.

The concept of universal acceleration with inertial forces "pinning" you down works fine as long as you are already on the ground, but not if you are supported above the ground.

If 100kg kid is on one side, he would be accelerated up with a force of 980N, with an inertial force resisting with same amount of force, 980N. The inertial force will "pin" him to the seat of the teeter totter to keep him from levitating, for lack of a better word, but it wont push the seat down to the ground if it is above ground.

980N up and 980N down means zero net force and zero net force means no movement for an stationary object.

Re: Center of gravity of objects on a flat earth
« Reply #15 on: August 26, 2020, 07:45:12 AM »
You need to ask yourself how 100kg-kid is "... supported above the ground".  He isn't suspended by levitation, he's only there because he's an integral part of a rigid ASSEMBLY comprising him, the 50kg-kid, and a beam, all attached to the planet by a pivot.  We assume that the beam itself is symmetrical and its mass is evenly distributed, so the COM of the ASSEMBLY is to the right of the pivot, because that's the end the 100kg-kid is sitting.   

If the planet accelerated up at 9.81 m/s/s, that force is going to be felt at the pivot of the ASSEMBLY.  The inertia of the ASSEMBLY acts at its COM which to the right of the upward accelerating force.  The 2 forces form a couple which rotates the ASSEMBLY clockwise until the planet hits 100kg-kid's butt, at which time the accelerating force is now distributed to both sides of the COM, via the pivot and his butt. 

Stop thinking of him in isolation.  He's just a part of something bigger. 

And have you noticed I've stopped calling him fat-kid. 


Re: Center of gravity of objects on a flat earth
« Reply #16 on: August 27, 2020, 09:43:39 PM »
You need to ask yourself how 100kg-kid is "... supported above the ground".  He isn't suspended by levitation, he's only there because he's an integral part of a rigid ASSEMBLY comprising him, the 50kg-kid, and a beam, all attached to the planet by a pivot.  We assume that the beam itself is symmetrical and its mass is evenly distributed, so the COM of the ASSEMBLY is to the right of the pivot, because that's the end the 100kg-kid is sitting.   

If the planet accelerated up at 9.81 m/s/s, that force is going to be felt at the pivot of the ASSEMBLY.  The inertia of the ASSEMBLY acts at its COM which to the right of the upward accelerating force.  The 2 forces form a couple which rotates the ASSEMBLY clockwise until the planet hits 100kg-kid's butt, at which time the accelerating force is now distributed to both sides of the COM, via the pivot and his butt. 

Stop thinking of him in isolation.  He's just a part of something bigger. 

And have you noticed I've stopped calling him fat-kid.

I didn't fully explain myself, my bad.

Assume we have a seesaw with the fulcrum in the middle with 100N downward force on the left and 50N downward force on the right (IOW, gravity).  It will rotate counter clockwise.

Now reverse the direction of the force (universal acceleration), with 100N on the left upward force and 50N on the right upward force.  It will rotate clockwise.  The heavier side will be suspended in the air, instead of falling to the ground.

My point was that inertia will slow down the rotation, maybe even stop it, but it won't reverse the rotation from clockwise to counterclockwise.

Re: Center of gravity of objects on a flat earth
« Reply #17 on: August 28, 2020, 07:03:34 AM »
You need to ask yourself how 100kg-kid is "... supported above the ground".  He isn't suspended by levitation, he's only there because he's an integral part of a rigid ASSEMBLY comprising him, the 50kg-kid, and a beam, all attached to the planet by a pivot.  We assume that the beam itself is symmetrical and its mass is evenly distributed, so the COM of the ASSEMBLY is to the right of the pivot, because that's the end the 100kg-kid is sitting.   

If the planet accelerated up at 9.81 m/s/s, that force is going to be felt at the pivot of the ASSEMBLY.  The inertia of the ASSEMBLY acts at its COM which to the right of the upward accelerating force.  The 2 forces form a couple which rotates the ASSEMBLY clockwise until the planet hits 100kg-kid's butt, at which time the accelerating force is now distributed to both sides of the COM, via the pivot and his butt. 

Stop thinking of him in isolation.  He's just a part of something bigger. 

And have you noticed I've stopped calling him fat-kid.

I didn't fully explain myself, my bad.

Assume we have a seesaw with the fulcrum in the middle with 100N downward force on the left and 50N downward force on the right (IOW, gravity).  It will rotate counter clockwise.

Now reverse the direction of the force (universal acceleration), with 100N on the left upward force and 50N on the right upward force.  It will rotate clockwise.  The heavier side will be suspended in the air, instead of falling to the ground.

My point was that inertia will slow down the rotation, maybe even stop it, but it won't reverse the rotation from clockwise to counterclockwise.

Even in a vacuum, the  acceleration is going to produce an upward force on the fulcrum through a normal force exerted by the ground  Since the force does not occur at the center of mass of the seesaw, a torque is exerted .This will cause the seesaw to rotate exactly as it would in a gravitational field,

It’s been 100 years since The Equivalence Principle was proposed.  If there was a hypothetical experiment that would have violated it, I think someone would have figured it out by now.

Edit:  Reading through the above threads, with the seesaw/teetertotter example, we need to consider torque.  Simply looking at the forces doesn't tell the whole story, because the seesaw system as a whole its no longer a rigid body.  As mentioned above, the degree to which the seesaw rotates, is going to depend not just on the mass balance at either end, but where the fulcrum is placed.  This is a separate issue from the CoM, as we can make a simplification that the mass of the board connecting the children and the  mass and that of the fulcrum are small in comparison to the weight of the two children.  But if we shift the position of the fulcrum, it changes the behavior without affecting the CoM or the forces involved. 
« Last Edit: August 28, 2020, 07:33:36 AM by Jeb Kermin »

Re: Center of gravity of objects on a flat earth
« Reply #18 on: August 28, 2020, 11:30:43 PM »
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Even in a vacuum, the  acceleration is going to produce an upward force on the fulcrum through a normal force exerted by the ground  Since the force does not occur at the center of mass of the seesaw, a torque is exerted .This will cause the seesaw to rotate exactly as it would in a gravitational field

Normal force is only as strong as the accelerating force.  They cancel each other out. Technically normal force causes acceleration, but since net force is zero, there is no motion.  If you've got an acceleratormeter n your phone, turn it on and place it on a table.  It will show acceleration equal to gravity.  Its measuring the normal force, but your phone isn't moving.

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It’s been 100 years since The Equivalence Principle was proposed.  If there was a hypothetical experiment that would have violated it, I think someone would have figured it out by now.

Equivalence principle only applies to freely falling objects, not to mention limited space and time.

Quote
Edit:  Reading through the above threads, with the seesaw/teetertotter example, we need to consider torque.  Simply looking at the forces doesn't tell the whole story, because the seesaw system as a whole its no longer a rigid body.  As mentioned above, the degree to which the seesaw rotates, is going to depend not just on the mass balance at either end, but where the fulcrum is placed.  This is a separate issue from the CoM, as we can make a simplification that the mass of the board connecting the children and the  mass and that of the fulcrum are small in comparison to the weight of the two children.  But if we shift the position of the fulcrum, it changes the behavior without affecting the CoM or the forces involved.

Direction of rotation is always in the direction of the strongest torque.  Torque is T=F*D.  If the fulcrum is in the middle and the distance is the same, rotation will be in the direction with the greatest force. With 75N of upward force on one side and 50N of upward force on the other side, rotation will up, towards the side with 75N.