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so from the 4000N longitudinal force you get

4000N longitudinal and 3000N lateral ???????

I think that you will find that the resultant
longitudinal forces will only be 1000N.

I didn't use a 4000N input. I used a 100kg mass bashing into it at 40m/s. But yea those were the average forces at the supports.

Go back to the FBD to see if it should be 1000N or not. No point just making up numbers because they feel good. Also no point making up a 'law of conservation of magnitude of force' when none exists. This mechanism has a mechanical advantage or disadvantage, so you can tune it to give whatever horizontal force you want. You can have 1000N if you want, but it won't detract from the 4000N pushing backwards.

100kg mass bashing into it at 40m/s.
thats 4000N isnt it?
at least it is if the mechanism stops
the 100kg mass in 1 second.

whats the name of the program you used?

the reason I say 1000N longitudinal
is because you say 3000N lateral.
you cant have a reaction larger that the action.

for every action there is a equal and opposite reaction.


the cosine of 45 degrees is 0.7071

4000N * 0.7071 = 2828.4N
this should be the resultant lateral force that the
slider feels.

of course you would divide that by 4
when using 4 legs , you can use 1 to get the total.

pretty close to your 3000N !

Yep, on average it is. It doesn't apply a constant 4000N.


LISA www.lisa-fet.com


The 'action' on the left spring is -1500N, the reaction applied by the spring is equal and opposite, +1500N. Similarly for the other spring.

Again, there's no reason to conserve magnitudes of forces. You can easily apply 4000N longitudinally and get 10,000N sideways. It's just mechanical advantage.


Yes, opposite, not sideways.

Can you go into more detail? You seem to be considering both legs together, as well as the input force. It's a bit confusing.

But I think the general idea of calculating components is right.

Hey by the way, do you tend to edit your messages, adding extra parts, and turning off the 'mark as edited' option? I'm not complaining, but I often find your messages are different after I've replied to them. Or maybe it's a bug in the forum?

yes , opposite and opposing.
thats why you cant have the full 4000N , its not opposite.

the forces start at a 45 degree angle.
and the angle decreases.

yes , its always a work in progress , you should
usualy wait a few minutes to reply.

Word games mean nothing in physics. FBDs aren't ambiguous. You really should use them. They make it clear what all the forces are.

if you can agree to the above , then I will add this
to my program to return the resultant forces using distances and time for each degree.

in the above none of the resultant forces are above the
4000N applied , so I trust this.

what I meant in saying you cant have the full 4000N
was that the longitudinal direction that you are claimimg that the 4000N force would be in , would not be 4000N at no time durring the cycle.

it will always be from 45 and decreasing as pictured above.

Yea those diagrams perfectly correct in themselves.

But it's not 4000N along the diagonal member. It's 4000N longitudinally at the input, so that direction change has to be calculated in a similar way.

As well it's split between the two diagonal members, so they only get half each.

I was just showing how the forces would direct through
the point that you used , of course the first angle
from the input point would also be at 45 degrees.

you didnt include that either did you?

yes I was intending to include 4 legs in the program
and using all elements involved , we can call the sliders
massless or we can give them a mass , also friction
in space would be the result of forces causing objects
to rub against each other , so there will be friction
should we use a friction coefficient also or can we
say that friction is negligible for now?

So if you can agree to this then I will put these
into the program Im building.

do you think you could help with a spring coefficient
to use?

I asked earlier but you didnt reply , what did you use
in the program you used?

Yea I agree with what you did and that the first angle would be the same. Just not with using 4000N on the diagonals.



I used a spring constant of 70 N/m for each spring.

Everything massless would be good. And everything frictionless. Those two things can always be made arbitrarily small in real life.

Im almost done with the program , but I have to say
its not looking good for your side , LOL.

well I am done with it mostly , the calculations are done
I just need to make a postable text file with it that
shows the results.

the 70N/m stops the mass to quickly using a 20 meter
stopping distance , it stopped in less than 1 second.

and in a 40 meter stopping distance it stops in 5 meters.

so I need to find a useable stopping distance and spring co efficient to use.

I want to make sure its all right before I post the
results anyway.

Remember I won't be able to trust your computer results any more than you trusted mine. You'll have to describe the calculations in such a way that I can easily check and reproduce them.

What are the equations? And why not do the static calcs anyway? We never resolved even the static case, so I don't expect any dynamic analysis to have much purpose at this stage.

good point

maybe I should include the equations in the readout that I make.

I havent determined the best arm lenght and its initial and
final angle yet , so when I do then I can include the
readout portion within the program.

Im using the spring co efficient as a resistive force.
and then I subtract that force from the 4000N before continueing
to the next set of calculations , so it should be pretty precise.

now the more sets of calculations that I allow the program to perform , the more precise the results are.

so it will be awhile , meanwhile I have recieved my new starter
for the hydrogen engine im building.
I purchased a old 15 hp engine frm the salvage yard and am
rebuilding it to use HHO only !!!

if it works well enought I will want to find a generator head
to attach to it , and then get my free energy.

it has a built in generator that charges batteries and powers lights and stuff on a riding lawn mower and this generator might just make all the HHO it needs to run.

Don't bother optimizing it. To prove me wrong you only have to show that either of these claims is wrong, even slightly:

1. The average of the total force applied to the pipe is identical to the average force applied by the mass.

2. For a 100kg 40m/s mass, the average force applied by the mass is 8000Ns / turnaround time.



It isn't a force. You have to multiply it by the amount of spring deflection to get the force.



Cool. I'm glad you're actually trying it. I just hope that when it fails you don't make some excuse about too many strokes or not fully using the intake vacuum or whatever.

I suppose you mean the average force applied to the pipe in the
longitudinal direction in #1 above.
because alot of the force is being applied in the lateral direction.

and the force that is being applied in the longitudinal direction
is being applied as a torque or twisting force that stresses the
pipe along its sides , therefore that force can be dampened also
resulting in even less longitudinal force.

in #2 above it is 4000N that is required to stop the mass in 1 second as the mass stops , however it seems that the springs do not compress enought to return the mass at 40 m/s in the next second , so the total will be less than 8000N.

alot of the force is lost through the angles and in the twisting torque through the sides of the pipe.

but dont worry Im taking everything into consideration.

maybe I should make another drawing or two to explain this better.

That would do too. I actually mean the entire total force. All lateral forces cancel out in the averaging operation, so only the longitudinal component contributes.


Equations not words! If that's true, I'll admit I'm wrong, and you'll be confident that your machine will fly.


OK. I still stand by 2nd point, except I realize I used the wrong sign convention. Should be a negative in there. I'll add a more general #3 to allow for your new situation:

2. The average force applied by the mass is
F = -m * (v_2-v_1) / turnaround time.
Where v_1 is the initial velocity (-ve for backwards), and v_2 is the final velocity. This allows it to come out slower than it went in.


I've also made some unstated assumptions in how the averaging is done. But I don't think it'll become important unless you have masses inside the device. The averages should both be made over the same time period and it should be one in which all forces occur, even if that's longer than the turnaround time.



You should show some equations or even code. Actually I'm amazed you're programming a dynamic response solver. Engineering students do that in their final year. You really should do a FBD or any kind of static analysis. If there's mistakes in that then surely any work on the program will be a waste.

Im doing this in vb6 , because its much easier , do you have vb6 or higher?

if you have higher I could probably upgrade the project
to fit in with your compiler.

or just copy n paste the code and you can figure it out.

I have to warn you though its not pretty and I use a few tricks
to speed things up a bit.

Oh ouch. I've had plenty of bad experiences with people trying to optimize VB6 code (don't do it!!!). I think I'll pass. At least until it's complete.