Recent content by Matt_B

here is the main problem with that theory... if you make it conical, or vary the pitch or even vary wire thickness, you are still going to reach a point where somewhere in the travel the force applied to continue the compression will have to increase.

What i am saying though, is that it is not hydraulic or pneumatic, nor electronic. It actually looks like a normal spring you would see in a car when you look at the suspension. The only difference is that it does not raise in resistance when deformation increases

From what i am reading on that hit list you provided what they are calling "constant-force springs" is a roll spring which is not what I am talking about, i am talking about a compression spring

Not really something I am working on at all, but a while ago I had heard that engineers for some of the race teams, mainly NASCAR have developed and put into use and compression spring that has a uniform force of the entire distance of deformation. To put it better, you have a suspension...

Wow, thank you DH, that helped explain a lot! So basically the only way that F=MA works is when force is in Newtons and mass is in Slugs and Acceleration is in m/s^2 ? This is making a lot more sense to me now

Actually let me ask this first My understanding is that if you take one object and drop it from a given height, then you make that object 2x as long and drop it from the same height, that even though the mass is 2x as much, the force at the moment of impact is identical, but the change is...

Ok thanks, that makes complete sense :) I am currently working on how to phrase my next major question... keep a look out for it! I think this one is a doosey

you misunderstood my question... if i am accelerating an object, and the line of acceleration is perdendicular to gravity. Lets say that the force of gravity is pulling down in the vertical direction. and i am using my spring to accelerate in the horizontal axis.

Ok, since Earth's gravity is the constant here... if you change location to where gravity is different, let's say the moon... is it safe to say that the acceleration of an object will also change when accelerating perpindicular to gravitational force?

Ok, so over my distance i am getting 16.54 ft/s ok, now i am really lost! LOL

Correct me if i am wrong, but recalling gravitational acceleration, i thought that mass did not come into play? It goes back to the jr high theory that in a perfect vacuum, if you drop and bowling ball and a feather at the same height at the same time, they will hit the ground at the same exact...

So what you are saying is that with my given information it would look something like this: 6=\frac{.0106}{32.174}a which breaks down to 6= .000329a would would end up being 18,237 ft/s^2 ? that in no way seems correct, i mean we are talking that after roughly 1 second the...

yeah i got 2.95 ft/s thank you a lot guys, you are a HUGE help! We really appreciate it!

is this correct now?

Ok, so then i am at .09 = (1/2)*566*t^2 which would then turn into .09= 283t^2 so result would be .000318=t^2 is that correct? do i then take the square root of the .000318? so i am getting time at .017833 then to get velocity i V=566*.017833 ending in 10.09 ft/s ? and i am...