# Why a smaller spring will build up less force?

Thus, we see that K is analogous to 1/C , which makes sense, because with a given current for a given time, a smaller capacitor will build up more voltage, whereas with a given velocity for a given time, a smaller spring will build up less force.

I don't understand why smaller spring will build up less force? I know that $F = K\int vdt$ and if $v$ is the same so $x$ is the same, and as $K$ is the same, so force is the same.

Dale
Mentor
2021 Award
A smaller spring has a smaller K

A smaller spring has a smaller K
Why, usually if we cut spring in half, then the constant doubles.

Chestermiller
Dale
Mentor
2021 Award
No, you misunderstand. In this case “smaller” is not referring to physical size. “Smaller” as it is used here directly means smaller K. Just like a “smaller” resistor means a lower resistance regardless of the fact that a lower resistance can be achieved by making it wider.

harmyder
sophiecentaur
Gold Member
"Smaller" could mean shorter, narrower coils or thinner wire. You would need to specify which meaning you are choosing before coming to a conclusion.

ZapperZ
Staff Emeritus
Why, usually if we cut spring in half, then the constant doubles.

As Dale has stated, this is not the physical size! I can have two springs of identical size, and yet, they will have different spring constant k. This clearly indicates that k depends on many other factors.

Zz.

When talking about capacitors in a circuit it's common to use "large" and "small" to refer to the capacitance. You put small capacitors (nF or low uF) near integrated circuits for decoupling and large capacitors (high uF or mF) on the output of a power supply.

Dale
sophiecentaur
Gold Member
When talking about capacitors in a circuit it's common to use "large" and "small" to refer to the capacitance. You put small capacitors (nF or low uF) near integrated circuits for decoupling and large capacitors (high uF or mF) on the output of a power supply.
There is not a simple correspondence here. When we say a 'small Capacitor' we imply a 'small Capacitance'. Physical size is secondary until way down the line in circuit design. There are more design features in a spring that are of instant relevance.
Why not just bite the bullet and describe the spring in ways that allow a proper discussion of how it will perform in this launcher? Spring constant is a good place to start of course but length, compressibility (coil spacing) etc, are all very relevant here.
The electrical analogue is trivial in comparison with this project. How many of us have used spring calculations in a construction as often as calculations of circuits with capacitors in them?

Just goes to show sometimes analogies do more harm than good.

RPinPA
Homework Helper
I think "a smaller spring" in this article has no physical meaning except "a spring with a smaller k". It is just reading the equation, not trying to think about the material the spring is made of. It is comparing the role of k with the role of 1/C in the analogous equations.

The language is unfortunate.

Dale
sophiecentaur
Gold Member
Just goes to show sometimes analogies do more harm than good.
Actually that goes to show that the analogy needs to be appropriate. There is a 'perfect' analogy where Force corresponds to PD and Charge corresponds to displacement. You can't just take two pairs of variables and expect them to exhibit the same rule.

CWatters
Homework Helper
Gold Member
By smaller spring they mean a weaker spring. They hadn't considered that some people might interpret smaller spring to mean similar in construction but shorter.

sophiecentaur
sophiecentaur