150 replies to this topic

[TripleHelix]

Do you mean moving the trigger (the blue rod that locks the chain in place) towards the end of the launch? If so, I thought I already made that clear. (In the video the trigger is also one red rod size towards the end of the launch, which gives the rubber band more space to stretch).

Oh, no, I don't mean moving anything, only making the long black rod in the tutorial able to be rotated from the train hitting it by adding onto it. Like the black rod Gumps' talking about:

The train doesn't trigger the launch in this tutorial. You see the black rod sticking out to the side? Rotate this rod and the trigger will let go of the chain.

[Jesperdude]

Good because it is pretty dark in my room and photoshop had difficulties popping the lighting on.

making the long black rod in the tutorial able to be rotated from the train hitting it by adding onto it. Like the black rod Gumps' talking about

I think that would be difficult, since the trigger has to stay up for a while, not for a second (like in my signature). But it would definitely be great.

My words exactly;)

[~stεεlspectrum~]

Do you have to push the car back so the chain dog is pressed against the chain link every time? Because on Stealth, I had to make a mechanism to push the car back as far as it would go. If the chain dog wasn't pushed all the way back when it launched, the front car came apart.

[Ace o Spades]

Holy shit man! Ace demands Jesper gets Elite status NOW!

[Jesperdude]

Holy shit man! Ace demands Jesper gets Elite status NOW!

And I demand you to be more careful with a Wii!
Thanks Ace! Good to see you again too.

~stεεlspectrum~:
I don't take any risks on breaking the chain dog. However, when it launches, I am forcing it quite hard, so forget what I just said:) I don't push it back always to make contact with the chain.
If you're afraid of breaking the chain dog while it launches, someone came up with an alternative chain dog. Which is basically a small gray connector with a tiny green rod, in the center of a blue rod which is connected by two small black connectors underneath the train. (I'm looking for pictures as we speak) But I can't guarantee it works now, since I haven't tested it yet!

[TripleHelix]

My words exactly;)

Haha, sorry man, completely forgot about that. But I'm pretty sure the locking mechanism can be modified so that... you know what, I'll stop.


But, anyways, I'll save you the trouble of looking for something similar to ^ that:

[Jesperdude]

Thanks for looking, TripleHelix. When I looked at that picture I saw an immediate problem.
And I was in the mood for a video so here's most of my reply:



Also, I have made a quick draft of photo's for the push mechanism tutorial. Now, I know what to shoot and what not. I will realize this as soon as possible (I think on Wednesday I have time).

[christian]

To get this automactic let's bring a HUGE rollercoaster motor. That'll work .

-MIBB)

[TripleHelix]

Haha, I didn't expect it to be launched... backwards. The chain dog I showed was actually supposed to be used near the back of the train to push it over the crest of a lift hill as a second chain dog, not the primary dog (it'd work good for an inverted coaster). That's why the picture I took is on the back car because that'd be the most likely place for it.


And the only problem with the moving chain dog (second launch) is that it creates friction against the middle tubing.

[coasterfanatic11]

would there be any way to get this thing automatic?
and would it work with VV?

love the tut, really easy instructions

[Loop De Loop]

Dude I can not describe how shocked I am at this. You are an idol, a GOD if you will....

Dude you better become elite or im going to be really pissed off.

[RightWayTrey]

I really love the tutorial, not to mention the launch.

If you used my switch thing, I think it would be pretty easy to get the motor to push down the trigger for several seconds.

[Jesperdude]

I really love the tutorial, not to mention the launch.

If you used my switch thing, I think it would be pretty easy to get the motor to push down the trigger for several seconds.

Flipping the switch be weight instead of muscular strength, huh?
If so, I think that's a great solution. But however, instead of your very complicated mechanism (which is absolutely amazingly profound! congrats btw) I'm now thinking of a tower next to the track (doesn't need to be very high I guess, where something weighted will be lifted automatically, and will be released at its highest point. And then, it will fall on some "push" mechanism.

So basically, installing a drop tower near the launch to make it work. I think that can be a good idea (and it will be something less complicated than your hi-tech launch-swifting mechanism)(I hope).

What do you guys think of it?

BTW: I added a space for updates in the first post of this thread.

[~stεεlspectrum~]

^ I think RightWayTrey was referring to his automatic motor switch mechanism. I may be wrong, but it sounds like you two are talking about different things.

[Jesperdude]

I may be wrong, but it sounds like you two are talking about different things.

I'm pointing out that I got inspired by his method: working with weight.
I am talking about a complete different thing after all.

[knexems]

Well, I made it(I geuss I will post pics), and it's amazing! I made some very minor adjustments like moving the wheel in the bottom layer 2 blue rods closer to the end of the launch. I also decided to do a quick little experiment.

For a five car train, the distance from the tip of the train to the end of the launch is about 4.25 feet.
The time it took for the train to reach the end was 0.44 seconds.
That means the launch propels the train at about 9.66 ft/sec. Yay!

[SwimZONE]

How many MPH though, that's the question.

[knexems]

Alright then, hmm lets see.... that would be about 0.1097595 mph.
EDIT:Nevermind, I screwed up. Look...V

[RCT2123]

^I am pretty sure it's 6.59 MPH...

9.66*60= 579.6 ft/min
579.6*60= 34776 ft/hr
34776/5280= 6.59 MPH

If you want solve for acceleration the equation is V=aT

V=Veloctiy
a=Acceleration
T=Time

So if V=aT, a=V/T

Acceleration is 21.95 ft/s/s

-Rct2123:cool:

[Jesperdude]

Acceleration is 21.95ft/s/s

And that's about 6.69036 m/s

And is that good for a launch with 5 cars? Now THAT'S the question;)
Good job knexems. Since you're the first one who (probably) followed the tutorial, is everything understandable?