Science a GoGo's Home Page Forums General Discussion General Science Discussion Forum Methane Clathrates in Gulf and Worldwide

Since when? If you mean the 2nd accelerator that stops it, then no, you are still turning it around. There's clearly no way you can bring it back to the starting point without turning it round.

Of course you could just stop it at the 1st turn. The pipe gets 4000N from the accelerator and -4000N from the stop. Is that what you're saying?




Drawing a spring laterally is misleading. As i said it must have some stiffness in the longitudinal direction otherwise it'll simply shear off and do nothing.

We can get to that later if you actually think it'll tell us anything useful.

Obviously if this is an actual machine then it can, in principal be transported to earth. And it can be weighed. I'm asking you what would happen if you did that. I explained this in my message, you're just being contrary for no reason other than to attempt to hide your mistake.

the below is more like what I had in mind to direct the
force laterly.

OK that's what I thought. But it's missing any support in the longitudinal direction. The whole thing will just fall off when the mass hits it. Sure it might absorb some of the impact, but you don't need any fancy sideways springs or linkages for that, just a big block of rubber or a single spring floating in space.

yes , it just stops , its force is distributed laterly as in the images above.

then the mass is accelerated by the springs into the accelerator.

I didnt draw what the mass that pushes the springs are sliding on.

the springs are attached to the pipe.
the arms push the sliding mass that compresses the springs.


I thought you could just imagine them sliding on a support.

think of a rod that passes through the springs and the masses.

OK. But you forgot some forces on your diagrams:

yes there will be some force placed in the (-) direction
but the brunt of the force will be directed outwards.

your arrows are too long , they should more clearly represent the force.

they should be much smaller.
and they should get smaller and smaller as the sliding mass
slides outward.

OK. Those 4 sliding supports together carry the entire load applied by the mass. All 4000N, and they apply that to the pipe, pushing it backwards.

There's no "law of conservation of magnitude of force". You can't convert a 4000N longitudinal force to 4 x 1000N lateral forces. Sure you can get 4 x 1000N lateral forces using that mechanism, or you can get 1,000,000N laterally if you want. Either way it doesn't reduce the 4000N longitudinal component.

If they do get smaller then this is an antigravity device.

But they don't. In total, they're always the same as the applied force regardless of increasing lateral forces.

sure it does , its done all the time.

picture the sliding masses as cars.

the cars have weight because they are sitting on the earth
so this same setup would work on the earth.

if a force is applied to the center the force will extend to push the 4 cars away from the center , the 4 cars roll on the ground.

they still have weight.

but the force is directed away from the center.
because of the mechanism the force is not directed downwards.

Yes. And they also each have 1/4 the downward force pushing them into the ground, additional to their own weight.

Come on, didn't you ever play with Meccanno as a kid? This is a bog-standard toggle mechanism. It's a common way of amplifying forces. Don't you think somebody would have noticed that it also does antigravity?

forces are directional , the force that is applied in a direction cannot change direction by itself , so the cars feel a outward lateral force.

they do not feel any additional force.

in fact the force that the earth feels because of the
cars weight decreases because the cars are moving.

if a car is riding down a street made up of weight scales
the car would appear to weigh less and less as it passes over the weight scales as the car accelerated down the street , because the force has changed direction.

this is why a car driven off a cliff at high speeds does not go straight down as soon as it clears the cliff it
falls at a angle due to the force of gravity acting on it.

the force is mostly in the direction that the car is traveling at.

you seemed to understand the outward forces when we
were using the turn , in fact you believed that a
4000N force became 12,566N because the forces were
inward due to the acceleration towards the center of the turn that the 100kg mass experienced because it turned, but now you seem to think that forces can not be
directed outward as it was in the turn.

you need to get your story straight.

I can tell your getting stumped on this one , because
your reaching into the insult bag to try and cover
your losses.



I never had any toys like that , just tools and wood
and metal , and such.

I used to want toys like that but they were too expensive.

and Im not sure where you get amplifying forces
from, the 4 springs dont amplify forces , they
only direct forces , I didnt include any mechanical advantages in the design.

heres one other point you seem to be overlooking
completely , even if only a portion of the force
is directed outward , then the final result would
be that the pipe would be capable of forward motion
each cycle and its displacement would continuously increase because thats the way forces work.

if you cant defeat the 4000N , then the pipe doesnt stop.
it only slows , then as the mass is tossed back into
the accelerator and slowed to a stop in the accelerator
by the acclerator the pipe accelerates more , then the accelerator accelerates the pipe even more by accelerating the mass again , it never stops, as long as the accelerator is switched on.

so unless you can figure out a real reason that the mass will direct 100% of its (-) force to the pipe
in the (-) direction , then the concept is 100% viable.

and it will work !!!

Let's not get side-tracked into movement.



It didn't 'become'. 8000N was simply the average longitudinal component of the radial force.



I'm doing that because you're frustrating me. You're also being dishonest. Why didn't you show the downward forces in your diagrams? All those arrows but you neglected to include, or even mention the very ones that I had claimed would be there.




Yep, your mechanism gives more mechanical advantage the further it's flattened down. Look up "toggle mechanism". With 4000N applied on top, the side forces can increase way beyond 1000N each. In theory, they become infinite when the linkages are horizontal. That's how vice-grips work.



Exactly all of the applied force is directed in the same direction it's applied in. There are _extra_ forces directed outward, but these don't take anything away from the longitudinal force.



If you still can't afford Meccanno, you can build this with biro springs and popsicle sticks. Test it by placing it on a scales with a weight on top. See that the weight weighs less with the device under it.

I'm getting tired of analysing all your inventions only to have you tell me my methods aren't valid. You havn't once shown that any of my methods have been faulty. I was about to do it using Newton's 1st law, but I expect you'll say that might be wrong.

How about you analyse this one? Be sure to include references for your methods/equations/etc.

"this is why a car driven off a cliff at high speeds does not go straight down as soon as it clears the cliff it
falls at a angle due to the force of gravity acting on it."

True, but not because the force of gravity is redirected or anything. A Car driving with fullspeed over the edge of a cliff will hit the ground at the exact same time as a car dropped from the same height.

A more practical example would be the bullet of a gun, if fired exactly parallel to the ground. The bullet will hit the ground just as soon as any other object falling from the same height.

Also a car driving on a flat street doesn't become lighter. The donward force of gravity is independent from any sideway forces.

Here is something you can try for yourself:
Get some weight (e.g. 1kg) and a string.
Put the weight on the middle of the string.
Attach one end of the string to a door/wall.
Now you can pull on the other end of the string.

The 1kg of the weight will always have a downward force of 9.8 N.

Nevertheless you can pull with all your strength, there will be always a slight tilt of the string. There is an exact relation between the tilting angle and strength you are pulling at, so that always exactly 9.8 N are applied in a vertical direction to support the weight.

its not the weight kallog that Im directing , its the force
of the mass times its acceleration.
Im sure you understand that but you cant find a way to get
force to fit into your naysay.

so you use a example of a static weight sitting on a scale.

of course the mass would WEIGH the same sitting on a mechanism such as the 4 springs , but that has nothing to do with what Im talking about.

Its really funny the way you wanted to use 8000N as a (-)
force in the turn , starting with a 4000N force.

then when the cards turn you no longer use outward force.

LOL.

and now you say that all the (-) force is still there
and its added to the force that has been sent outwards ,
how do you suppose you can apply a force in 1 direction
then divide that force in 4 directions and still
have the same amount of force in the (-) direction?

what you are sudgesting is free energy.

Im saying the 4000N is divided up in the 4 lateral directions and in the (-) direction.


because if the 4000N force can be reclaimed in the
(-) direction , then all the other force is free !!!

the trouble with that is that it cant be done.

just show me how you do that with a force diagram.

shut up momos none of your replies had any meaning to what I
was saying.

your replies are out of context , stay out of this.





you think Im wrong put your weight scale in your yard
run over it and see what the thing says.
LOL



gravity is the only reason the car or a bullet or anything else falls , you think Im wrong , hold your computer out the window and drop it?

dont throw it downwards or horizontaly , just drop it.
or you can throw it upwards to make it last longer.

I actually don't understand that at all. Until now I thought you were only redirecting forces. What exactly do you mean by "force of the mass times its acceleration"? Is that a cross product? Maybe you mean "the magnitude of the force times the magnitude of the accelearation"? That's a scalar quantity which doesn't have any direction to redirect. Can you be clearer?