Tungsten Canopy

Feb 28, 2015
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I bought 4 tungsten canopies at $20 apiece before I found out that you guys are generally using 1/4 cubes. Is the reason that people stay away from the canopy because of the loss of speed due to :
1) aerodynamics (as it sits above the body),
2) it doesn't sit back far enough beyond the rear axle or
3) some other reason that I haven't thought of ?

I'm sure someone has run a test to see how much difference using the canopy costs time-wise versus the cubes. What time differences did you find?

Do I need to ebay those canopies?
 
#1 and #2 correct, IMO.

I would also add that the higher the weight is, the less stable the car. And, there is no adjust-ability to "move" the weight as the car needs it.

Just some quick thoughts. I'm sure there are more reasons...
 
My son and I used that once...the car looked cooler than a fan, but that car was the most unstable car I have ever built. It litterally would fly off the track every time in random spots. It was horrible, but it sure did look cool. There is a pic of the car in the pic section, it was the green lantern car.
 
we have used one the last three years for my son's Cub Scout race. Our cars are dead solid stable and have won the pack every year. Won district the last 2. Looks cool, too.

Doubt it would work here, though
 
OPA, Txchemist can probably speak more facts to this, but I believe I remember reading about the lower the weight is in a car (Center Of Gravity), the more speed you can get.
 
I guess what I am hearing is that I can use the tungsten canopies for scout racing, but it would be a detriment to race in the league with them. Thanks, guys, I appreciate the advise.

BBU
 
For what it is worth, my sons ran a couple of cars with canopies in their pack and district races and both did quite well. The cars were stable in all their races and were among the fastest in the field.

If you have them, my vote would be to use them!

Greg
 
ODD said:
Txchemist can probably speak more facts to this, but I believe I remember reading about the lower the weight is in a car (Center Of Gravity), the more speed you can get.

I lack TxChemist's (or 5Kids', for that matter) graphic ability. But, I can do the math...

Say you have two cars. Car1 has the weight 3" up, at the very back of the car. Car2 has its weight at 3/8" high. Everything else is equal. They are the same length. No friction (aerodynamic or otherwise).

Further, the cars start from a height of 48", plus the further height of their weights on the hill.

So, the start height of Car1 is: 48" + cos(27.1)*3" = 48" + 2.67" = 50.67"
The start height of Car2: 48" + cos(27.1)*0.375" = 48" + 0.33" = 48.33"

Things are looking good for Car1, right?

Now, how far did their CoMs fall?
The fall height of Car1: 50.67" - 3" = 47.67" (Remember, the CoM is 3" high in the car)
The fall height of Car2: 48.33" - 0.375" = 47.95"

Uh oh.

Do some maths to figure out speeds at the bottom of the hill using v = SQRT(2*g*d) <- (g = 32'/s^2)

v1 = 15.94 ft/s
v2 = 15.99 ft/s

Now, I don't want to go into the minutia of calculating the whole acceleration from zero, bottom of the hill, etc. Let's just say that both cars hit the bottom of the hill at the same time and now have 32' of flat to run (technically, Car2 has a nose on Car1 now, so has less distance to run...).

t1 = 32' / 15.94 ft/s = 2.0075s
t2 = 32' / 15.99 ft/s = 2.0013s

Car2 wins the race by 0.0062s solely by virtue of vertical weight placement.

---------

TL;DR: Lower weight placement is faster.
 
Crash,

All I can say is WOW. I do love the math. Just to make sure I understand your formula:

48" + cos(27.1)*3" = 48" + 2.67" = 50.67"

48" = the height of the start point
27.1 = the angle of the track (?)
3" = the height of the weight on the car (?)
2.67" = the distance from the front of the car to the COM (?)
50.67" = the total length of the vertical drop for the car as measured from the COM (?)

BTW, I am loving this site. Thanks in advance for sharing this information.

Greg
 
67 Rally Sport said:
Just to make sure I understand your formula:

48" + cos(27.1)*3" = 48" + 2.67" = 50.67"

48" = the height of the start point
27.1 = the angle of the track (?)
3" = the height of the weight on the car (?)
2.67" = the distance from the front of the car to the COM (?)
50.67" = the total length of the vertical drop for the car as measured from the COM (?)

48": If the body of the car was lying on the track, that is the height of the end of the car from the floor.
27.1: The angle of the start slope of the NPWDRL track.
3" (& 3/8"): How high the center-of-mass (CoM) is when the car sits level.
2.67" (& 0.33"): With the car sitting on the 27.1° slope, that is how much higher, vertically, the weight is from the end of the track.
50.67" (& 48.33"): How high from the floor the weight is at the start.

Hopefully, everything else was clear enough. /images/boards/smilies/smile.gif

0313151444a.jpg
 
Thanks, I think I have a good visual image of what you are describing. I have to say, this really brings out the math nerd in me! Now I want to dig up the formulas to calculate acceleration from zero. Hopefully, I can get my kids interested in this...

Greg
 
i was afraid you were going to say that... It's been a long time since I have done any calculus. The good news is that my wife was REALLY good at calculus. The bad news is she may have already had enough of me and my growing obsession with Pinewood derby racing!

Greg
 
I'm not the sharpest crayon in the box, so I'm hoping you can help me understand why a weight placed higher on a car falls further than a weight placed lower? If you trace out the path each center of mass takes down the track, wouldn't the paths be parallel i.e. Same magnitude of fall?