Oh no I had a falling mass launcher and for state I converted it over to a spring launcher. It's physically not possible for a falling mass launcher to hit that speed.

So a falling mass launcher will always have a weight that drops a maximum of one meter. The max speed of this weight is V^2=2g(1m), so V=sqrt(19.6), or 4.4m/s. So the mass will only ever hit 4.4m/s, and in fact if you add a scrambler to that mass it slows down, since the scrambler's not falling to add energy to the system. Any attempt to work with a block/tackle system to move the scrambler faster than the mass also results in dividing the height the mass can fall, which slows it down proportional to the square root of the change in height, so it's not ever made up to push the vehicle much faster. Or something like that. I'm half watching football and half trying to reason this out so correct me if I'm wrong :P

I used a push rod before switching to the Falling Mass/Spring launcher, mine wasn't too accurate. I've seen some that are, but they also have the speed challenges of the falling mass launcher. Frankly, I think push rod launchers are faster than falling mass (me and another kid in my state run scrambler one after the other, he with falling mass and me with push rod, mine was significantly fast, and he even had the lighter vehicle). Honestly a spring launcher is the way to go if you can get it to work well (I think ours was working very well at state, we did a conservative run and then the ETV got caught on a string, hadn't seen it happen before...).

Also, yes, ramps are clearly disallowed, but other integrated designs are still legal I believe.

'If you're the smartest person in the room, you're in the wrong room' - Unknown

I have a question regarding the dimensions of the scrambler launcher now that the bonus requires it to be angled. Does the measurement base themselves on the length and width of the launcher itself or the track? Like would a scrambler launcher in its ready to run configuration be measured with the tape measure perpendicular/parallel to the track or the launcher?

"Any sufficiently advanced technology is indistinguishable from magic." - Arthur C. Clarke

A Person wrote:I have a question regarding the dimensions of the scrambler launcher now that the bonus requires it to be angled. Does the measurement base themselves on the length and width of the launcher itself or the track? Like would a scrambler launcher in its ready to run configuration be measured with the tape measure perpendicular/parallel to the track or the launcher?

I would assume that it is measured on the length and width of the launcher itself. Submit a rules clarification in October.

"One of the ways that I believe people express their appreciation to the rest of humanity is to make something wonderful and put it out there."

A Person wrote:I have a question regarding the dimensions of the scrambler launcher now that the bonus requires it to be angled. Does the measurement base themselves on the length and width of the launcher itself or the track? Like would a scrambler launcher in its ready to run configuration be measured with the tape measure perpendicular/parallel to the track or the launcher?

Can you explain why you think the launcher needs to be angled? In order to get around the can?

A Person wrote:I have a question regarding the dimensions of the scrambler launcher now that the bonus requires it to be angled. Does the measurement base themselves on the length and width of the launcher itself or the track? Like would a scrambler launcher in its ready to run configuration be measured with the tape measure perpendicular/parallel to the track or the launcher?

Can you explain why you think the launcher needs to be angled? In order to get around the can?

Yeah, to get around the can you'd have to start with the launcher angled. Since the vehicle can only turn in one arc (to keep construction simple) it'll have to start aimed to the side of the track, and it'll turn to be running straight when it passes the bucket, and then past the bucket it'll turn towards the center to hit the center of the target wall.

'If you're the smartest person in the room, you're in the wrong room' - Unknown

A Person wrote:I have a question regarding the dimensions of the scrambler launcher now that the bonus requires it to be angled. Does the measurement base themselves on the length and width of the launcher itself or the track? Like would a scrambler launcher in its ready to run configuration be measured with the tape measure perpendicular/parallel to the track or the launcher?

Can you explain why you think the launcher needs to be angled? In order to get around the can?

Yeah, to get around the can you'd have to start with the launcher angled. Since the vehicle can only turn in one arc (to keep construction simple) it'll have to start aimed to the side of the track, and it'll turn to be running straight when it passes the bucket, and then past the bucket it'll turn towards the center to hit the center of the target wall.

Yup. And if the launcher was not angled, you would have to increase the width of the launcher to accommodate the tilted vehicle.It's much more efficient to just angle the scrambler and launcher themselves, rather than just the scrambler and not the launcher.

I think I said that in an understandable way.

"Any sufficiently advanced technology is indistinguishable from magic." - Arthur C. Clarke

I was just reading the third page of this forum and saw that people were saying that it isn't possible to go 1.2 seconds with a simple falling mass system, but last year we were able to get that time consistently. I was a sophomore then and the seniors made this really good one and when the state team was decided, I got paired with one of the seniors. We made a few adjustments but still managed to get that time.

Scioly isn't a club, or an organization. It is a lifestyle.
~Munster High School Science Olympiad Captain 2016~

Well, umm... For a simple falling mass system, like this

(from the wiki)
it's physically impossible to do that time unless the mass falls from more than 2.27m. (I'm not saying that wasn't your time, just that your system probably wasn't 'simple').

The total energy available to the system at launch is the potential energy in the falling mass.
[math]E_t = m_f g h[/math] Where [math]E_t[/math] is total energy and [math]m_f[/math] is the mass of the falling mass
The total system energy at the release of the vehicle is
[math]E_f = {\frac{1}{2} m_f v^2 + \frac{1}{2} m_v v^2[/math]
since the velocity of the mass and vehicle are the same, simply
[math]E_f = \frac{1}{2} (m_f + m_v) v^2[/math]
Now, digressing. The timed portion of the run is 8m, since the time in this case is 1.2s, that gives us a velocity of
[math]V_f = \frac{8m}{1.2s} \approx 6.67 \frac{m}{s}[/math]
We can use that for v in our energy equation, yielding
[math]E_f = \frac{1}{2} (m_f + m_v)6.67^2[/math]
or simply
[math]E_f \approx 22.24(m_f + m_v)[/math]
Since [math]E_t[/math] and [math]E_f[/math] must be equal (assuming a frictionless system) by conversation of energy, we get
[math]m_f g h = 22.24(m_f+m_v)[/math]
or, since g is a known value
[math]\frac{9.8}{22.24} m_f h = m_f + m_v[/math]
Let's divide by [math]m_f[/math], split the right hand fraction into two, and simplify
[math]0.44 h = 1 + \frac{m_v}{m_f}[/math]
Solving for the height in terms of the ratio of masses yields
[math]h = \frac{1+r_m}{.44}[/math] where [math]r_m[/math] is the ratio of masses [math]\frac{m_v}{m_f}[/math]
Since our ratio of masses must, in all cases, be positive (there's no such thing as a negative mass in this case), we substitute [math]r_m = 0[/math], knowing that all larger values will yield a greater height.
[math]h = \frac{1+0}{.44}[/math]
which gives us the minimum possible drop height for a simple falling mass system to reach the speed required to do an 8m run in 1.2s
[math]h \approx 2.27[/math]
which is obviously not a legal height for a scrambler.
We can further check ourselves, assuming a vehicle weight of 0, the system is the same as simply dropping a mass from a height, which is basic kinematics.
Any mass accelerates at
[math]g=9.8\frac{m}{s^2}[/math]
We can find the time is needs to accelerate to reach
[math]V_f =6.67\frac{m}{s}[/math]
using
[math]v_f = v_i + at[/math]
simply
[math]6.67 = 0 + 9.8t[/math] therefore [math]t \approx 0.68s[/math]
since we have a time, we can now find a distance required to drop
[math]x = \frac{1}{2} a t^2 + v_it[/math]
simply
[math]x = 4.9(.68)^2[/math]
or
[math]x = 2.27m[/math]
The same answer we got using conservation of energy.

'If you're the smartest person in the room, you're in the wrong room' - Unknown