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Plz help (horizontal spring launcher)


halo-recon

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Hey

 

I have to build a spring launcher to hit a target (7-8m away) with the spring by extending it. If i am given the launch angle and the distance d to a target (about 8m), how do i find the extension x that the spring must be pulled back i order to hit the target. (ignoring air resistance).

So I have to develop a general equation for determining x (the extension) in terms of the given variables (theta) and d (distance from launcher to target). Assuming that the launcher and target are both on the floor.

I another target is above the floor and only d and height (y) is given, how do i find the launch angle. Im guessing i have to find them by developing a equation.

 

PLZ HELP OUT!!!!! I will really appreciate it.

 

Launcher is simple consisting of a ruler a protractor and a mechanism to launch the spring...

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if you have said spring launcher the imple answer is to measure distance traveled at a few known extensions. This will give you the points on your graph, which will in turn allow you derive a quadratic equation to get your answer.

 

Without a table of values you're S.O.L.

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:googleit: get said table, and I'm shure I or someone else around here can help you, but this is something where you need DATA to calculate from.

 

there are too many possible types of sprigs, with too many different compression/extension ratios to know how any particular one will behave without knowing the specifics of that particular spring in action.

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Hello Halo, welcome to Hypography.

 

As I understand this physics lab assignment, you are to devise an equation describing a spring projectile hitting some target given the angle of the launcher and the distance the spring is pulled back. The teacher won’t tell you the distance of the target or the angle of the launcher until the day of the lab, so you need a general purpose equation.

 

As Sanctus points out, the first step is to find the elastic (spring) constant of the spring. It seems the teacher would either tell you this or make you find it yourself. But, it’s going to be needed one way or another. You can find it yourself using Hook’s law: F = Kx. In this equation, F is force in Newtons, K is the constant you are looking for (in units of Newtons per meter), and x is the distance the spring is displaced from rest in meters.

 

Rearranging Hook’s law, you get K=F/x. So, all you have to do is apply a known force on the spring and measure the displacement. Divide the two (F/x) and you get the constant K.

 

The constant K is going to help because we are going to need to know the velocity of the spring as it leaves the launcher. This is possible by using the potential energy stored in a displaced spring:

[math]E_P = 1/2Kx^2[/math]

where K is the constant you found above. We then use the kinetic energy of a mass at velocity,

[math]E_K = 1/2mv^2[/math]

The potential energy of your spring will be converted to kinetic energy when it is released and becomes a projectile meaning potential energy will equal kinetic energy ([math]E_P = E_K[/math]).

[math]1/2Kx^2 = 1/2mv^2[/math]

Solving for velocity using a little algebra, we get:

[math]v = x\sqrt{K/m}[/math]

We now have the velocity of the spring as a function of x (how far you pull it back). K and m on the right hand side of our equation are constants and known, so we basically have v = x (times some number).

 

The next equation needed is the one Turtle linked to above, the range formula,

[math]R = \frac{v^2 \ sin(2 \theta)}{g}[/math]

Here R is the distance to the target you want to hit, v is the velocity of the projectile as it leaves the launcher, theta is the angle of the launcher, and g is the acceleration of gravity (if you used meters in your calculations above, you should again use meters for the acceleration of gravity, 9.8 m/s^2, and meters for the range, R). If we substitute our previous velocity function into this equation, we get,

[math]R = \frac{(x \sqrt{K/m})^2 \ sin(2 \theta)}{g}[/math]

solving for x,

[math]x = \left(\frac{Rmg}{Ksin(2 \theta)} \right)^{1/2}[/math]

Assuming I’ve not made any algebraic mistakes, this now tells you the distance to pull back the spring (x) given the distance to target ® the three measured constants mass, acceleration of gravity, and spring constant (m, g, and K), and a chosen angle theta ([math]\theta[/math]).

 

To calculate hitting a target off the ground you would do the same as above, but rather than using the range equation from Turtle’s link, you would use the equation under the heading “Will it clear the fence?”.

 

Good luck!

 

~modest

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Thx alot. this really helped. i knew that i had to find the spring constant. we are to calculate k and find the mass of the spring. and when we have to launch the spring we will be given the launch angle and how far the distance is located 4-8m. Then we have to use an equation to find the extension of the spring. i get it now that i need a table of values in order to calculate it. when we are done with the calculations, then we will test them in our launch to see if we calculated the extension right. and now i have a general idea on how to approach this.

 

THANKS!!

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  • 2 months later...
hi, i was just wondering instead of relying on the spring; you could rely on the angle?

if that makes sense

 

Welcome to Hypography Meleapen,

 

Yes, absolutely, you can adjust the angel rather than the distance of the spring.

 

How far the spring flies depends on 2 things: how far you pull back the spring and what angel you set the launcher. All other things being equal, those become your two variables. The equation I gave above is for solving how far to pull the spring back given a certain angle:

[math]x = \sqrt{\frac{Rmg}{Ksin(2 \theta)}}[/math]

These variables are:

  • x = how far you pull the spring back in meters
  • R = range: how far you want the spring to go in meters
  • m = mass of the spring in Kg
  • g = acceleration of gravity 9.8 m/s/s
  • K = spring constant in Newtons / meter
  • [math]\theta[/math] = angle of launcher from horizontal

 

To use this equation you would choose an angle such as 45 degrees (or 0.785 radians if your calculator uses radians). You then plug in x, R, m, g, k, and [math]\theta[/math] (theta is the angle you chose—45º) and that will give you x (how far to pull the spring back).

 

What you're asking about can be accomplished by solving for theta ([math]\theta[/math], the angle). When we do this, it will let us pick an an arbitrary distance to pull the spring back and solve for the angle. You'd just have to be sure the distance you pick to pull the spring back is enough to accomplish the distnace.

 

If we solve for theta, we get:

 

[math]\theta = \frac{1}{2} arcsin \left(\frac{Rmg}{Kx^2} \right)[/math]

The variables are the same as before. The only thing that's new is the function "arcsin". This is something your hand-held calculator might not do. There's a Java application that will do arcsin here. You can also use Excel with the function "asin(...)".

 

If trigonometric functions, degrees/radians, or anything else I've made mention of gives you trouble, let us know. We can work through this step-by-step if you need. Also, If you just want us to check your calculations once you know all the constants, that will be no trouble.

 

Let us know how it goes, and Good Luck,

 

~modest

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I haven't studied Trig in 16 years. Can you help me out by showing how you went from sin to arcsin, please? :)

 

[math]\sin(\theta )= x \Leftrightarrow \theta=arcsin(x)[/math]

So in the equation in modest's post you just isolate [math]\sin(2\theta ) [/math] and then

[math]2\theta [/math] is just the arcsin of the term on the right.

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thanks so much!

just wone more question

say i make a catapult right?

So how would that be possible to calculate how far back youd stretch the springs to release it?

 

 

 

the catapult would be like, a base of wood, to pieces of would on the base (perpendicular to the base, but parallel to eachther); rod going through it; a stationary arm attached to the dowel (where the projectile would be placed)' then springs would be attached from the back for the parallel blocks to the back of the arm so when you pull back the arm the spring pulls, and when you let go the spring creates the force to let the ball go flying

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Well... Hummm... To tell you the truth, the derivation we did in post #7 for this equation is very situation specific. It assumes that the spring being pulled back will be the projectile—that no other forces or masses besides the spring will be acting on any part of the system.

 

What you're describing with a catapult and a ball is a whole different ball game. The physics would be very different. With a system that complex you will probably need to determine the velocity of the projectile through some method other than we did in this thread.

 

You might be able to fire a test shot and time the flight. With that data it would be possible to find the initial velocity which would then allow you to describe the physics of the system. This link might be helpful:

and this section in particular:

Also, the links in Turtle's post earlier in the thread might be helpful. But, I'm afraid without a detailed schematic of the catapult with the mass of each moving part... It just wouldn't be possible to create a general purpose equation :) ... unless another member has an idea I'm missing...

 

~modest

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