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Yah, but Siege said he's gonna check back in about half a year

Could someone give me a better idea of the pros of a spring launcher? It seems to me that the more gradual acceleration of a regular cable-counterweight launcher would be better than abrupt acceleration.

Hm..without doing any math..:

Imagine a much larger mass limit. say, 50kg.
If you use the direct cable-counterweight pulley system, your car will get pulled by the falling mass, accelerating at g m/s^2

If you instead, loaded the 50kg falling mass's energy into a huge spring, then shot the car, it would go much much much faster than just accelerating at g for 1meter.

Obviously, our mass falling, can only be 2kg.
So, doing it either way, can still give you relatively good results; thus why so few people try to make(accomplish and perfect) spring-launched scramblers

Well I'd really like if someone could show me some math
Because isn't there a certain amount of energy in a 2kg mass falling one meter?
And transferring it into a spring doesn't change the amount of energy.
So when you transfer that energy into kinetic energy in the car, shouldn't the same speed be achieved because the same amount of energy (excluding inefficiency in conversion) is used?

Someone excuse me if my physics is poor. The energy in a spring isn't quite the same...you see, there are different forms of potential energy and how they can be stored.
Potential energy in a spring = 1/2*k*x^2, where k is a spring constant and x is the displacement or distance essentially a spring goes
Potential energy in an object placed at a height = mgh, mass*acceleration due to gravity on Earth*height

If we make these equal kinetic energy we can show what happens when potential energy is transferred completely into kinetic (1/2mv^2), which can help us find v, or velocity.

With a falling object: mgh=1/2mv^2 turns to v=sqrt(2gh)
With a spring (and I'm assuming you're only using the falling mass to activate the spring, not to speed up the car in this case...hopefully that's right): 1/2kx^2=1/2mv^2 turns to v=sqrt(kx^2/m)

I think if you plug in numbers it shows that the spring's efficiency is difficult to get high unless you make it VERY well? At least, this is just my stream of consciousness of physics going here. I'm also not bothering with momentum, friction, etc, so hopefully this isn't reallyyyy bad.

syo_astro wrote:Someone excuse me if my physics is poor. The energy in a spring isn't quite the same...you see, there are different forms of potential energy and how they can be stored.
Potential energy in a spring = 1/2*k*x^2, where k is a spring constant and x is the displacement or distance essentially a spring goes
Potential energy in an object placed at a height = mgh, mass*acceleration due to gravity on Earth*height

If we make these equal kinetic energy we can show what happens when potential energy is transferred completely into kinetic (1/2mv^2), which can help us find v, or velocity.

With a falling object: mgh=1/2mv^2 turns to v=sqrt(2gh)
With a spring (and I'm assuming you're only using the falling mass to activate the spring, not to speed up the car in this case...hopefully that's right): 1/2kx^2=1/2mv^2 turns to v=sqrt(kx^2/m)

I think if you plug in numbers it shows that the spring's efficiency is difficult to get high unless you make it VERY well? At least, this is just my stream of consciousness of physics going here. I'm also not bothering with momentum, friction, etc, so hopefully this isn't reallyyyy bad.

I think the easiest way to look at the problem is to look at how the potential energy is being used in both scenarios. With gravity vehicle you do indeed have your potential energy (mgh(cm)) being converted into .5mv^2. This mass here is constant. All the mass that you have lifted needs to be translated to kinetic energy. You also lose energy to other sources like heat, noise, rotational inertia, and etc but this can all be made fairly small when you make your vehicle.
The important concept is that your mass isn't the same when you use a spring. The mass you drop (mgh(cm)) is converted into stored potential energy of the spring (.5kx^2) while you lose a lot of energy to the mass moving at the end because the mass still needs to move to get the spring in position (or at least it won't be perfect). Then you take that stored energy and convert it back into kinetic energy. But now your no longer dealing with m (mass of your weight), your dealing with the mass of your vehicle. So now if the mass of your vehicle is 4 times lighter than the mass that is dropped you can get a lot more speed. Although you lose a lot of energy to the spring (it's mass moving, it's beta coefficient of damping, etc etc). But if done correctly, you can theoretically go root of the mass's ratio faster. Hopefully I was able to help clear a few things up?

Ah I see, that's helpful, thank you. I think I understand.

From the perspective I was using before I think I see the flaw in my logic- all the energy of the falling mass isn't added to the car, because when the weight hits the floor it still has kinetic energy that the car never gets.
(I believe whatever energy the mass has has divided by the energy of the car is equal to the ratio of their masses?)

I was just wondering what everyone else was using to propel there car.

My team mate and I were are using a launching system.

We're using a launching system too!

What's powering yours? A launching system is a pretty general term... Ours is a falling mass. No fancy springs or pulleys or anything.

ours is the same thing just falling mass, two pulleys, and a string