If you were to go to diablo every day that it was open (not including holiday add-ons) you would be able to ride 105 days out of the year! Can you tell im really bored..

Oh and if you were a dumbass and didn't get a season pass to ride the 105 days, you would spend: $3,570 edit 1 big Gatorade a day $3 ea. $315 $3,570 +$315 -------- #3,885

the average MTB tire has a circf. of 81.64in a mile = 63,360 in. If you did do 2,100mi/season your wheel would have rotated 4,097,756 times... time for new hub bearings.

You know, some of that was actually interesting. How about total vertical/rotational distance covered by the rear axle as the suspension moves, assume average of 6 inches of movement 300 times a run.

yea then calculate how much the centrifugal wheel force would be on that if you pedaled let's say 150 full rotation crank spins per run. now that should give you plenty of time to kill to figure it out and don't just type some random numbers now, kill some time and develop some brain cells for real.

Well in 1 run your suspension would cycle 1800in. (really 3600in. including compression and rebound) So in a day (25 runs) 45,000 inches or 3,750 ft. (90,000" or 7500' with comp/rebound.) or... in a full season of 105 days... 4,725,000in. or 393,750ft. in a season 9,450,000" / 787,500'comp/reb.)

Energy forces need much more information than that... Raidus (meters) = .3302 Velocity (meter/sec) = depends on speed Velocity (kilometer/hour) = same as above Mass (kg) = depends on rider RPM (revolutions/minute) = depends on speed Actually the amount of crank revolutions per run would have nothing to do with the formula for centrifugal force. But if you want an answer.. if: Raidus (meters) = .3302 Velocity (meter/sec) = 8.9408 (20mph) Velocity (kilometer/hour) = 32.18688 Mass (kg) = 4.5359237 (10 lbs. rotational wheel mass) RPM (revolutions/minute) = ok explination on this one... Circf. of mtb wheel 81.64" 63,360in./mile at 20 mph your wheel will cover 776.1 rev./mile which means your wheel will cover .333 (1/3rd) of a mile. That means your wheel is spinning at 258.4413 RPM then: Centrifugal force (in newtons) = 1098.0988376284552 or 246.862439953226 pound-force

Wow.. I just realized, no wonder the shock companies recommend an overhaul after 20 hrs. with aggressive riding. 45,000 inches a day... just think of how much energy/friction/heat we create on that poor shock.

I only took one math course in college and it was easier than middle school algebra, wait it was middle school algebra, but I was 20 when I took it, oh well I had a buddy who spoek with the Sram tech guys in whistler about his World Cup boxxer and they were suggesting rebuilds every 4 hours

is that calculated using a 2.4 tire or 2.5 tire? i wanna know exactly what resistance measuring you used so i can calculate which tires are better. hahahaha i was just kidding on the centrifugal force, just giving you sarcastic challenges!

This is kinda interesting The physics of mountain biking No real world numbers or anything, just references to physics. haha found this too... its the statistics for nsmb.com post activity. and the stats for whistler/blackcomb mountain and a few funny quotes from mtbmilitia -If you can’t make it up the hill, it’s not because your bike sucks it’s because you’re fat & lazy. -Tube Math: tubes needed for any given ride = the number of tubes you have + one. -The more you hear the term "freeride" the more it will cost you to ride. -When someone says the trail you’re on is a deer-path, remind them that deer don’t have wheels. -Trip math: Number of miles actually ridden +4 ½. -Just shut-up and ride.

no prob.. broskee btw... panaracer fire XC's are the best tire, there is no mathematics for that. They have stripes to prove thier holyness and gnarlyness.

btw... noescape, nice work.. I tried to get into screenprinting. I dont have the patience for that $hit. Now I just do my basics with ThermoFlex

Im going to figure out the force requred by the gondola using: 2 riders per cart (170lbs. each) 40 carts per side gondola distance slope of the mountain 6 minutes to get to the top

Force as in torque at the drive wheels? You might also want to make it harder for yourself and figure out the force used in the detachment and moving system off the cable. Also, time is irrelevant in this case, because force has to do with just moving it. If you wanted to find the energy then time would be useful. This'll just be a lot of potential energy, plus it cancels out with the chairs moving down, so technically all you have to deal with is rider weight. Good luck!

Yup.. I realized last night I wanted to figure out the energy used. And I know the kinetic energy going up turns into potential energy at the top, then returns with negative kinetic energy, so the weight of the cars, cable, ect. cancles out.

Im going to say there is 80 total carts. The mass of the carts cancel out. The degree of the slope is 32º Average rider = 175lbs. X2 = 300lbs. Average bike weighs 35lbs. X2 = 75lbs. Total mass on cart = 375 lbs. Time to get to the top = 6 minutes. Distance to get to the top is 1962.56 That means the cart travels at 5.5ft/sec or 3.75mph The rate of change is 1.89 ft^/ft> so it moves at 8.75ft/sec. Now, F=MV or: F=(375)(8.75)=3281.25ft-lbs=.0012357716kWh That is the energy consumed to move one set or riders up on the lift. Let's say there is an average of 100 riders/day @20 runs so 2000 trips/day Also the lift runs for 420min./day which is 35 revolutions/day that would mean there is 57.14 riders on the lift at any given time. so 29 carts full at one time 29X.0012357716kWh=.0358353kWhX35revs=1.26kWh/day and... 1.26kWh/day X average rate of $.10/kWh = $12.60 to run the lift one day. Run 105 days in the season $1323.00 This does not account for wind or friction. And almost everything in this equation is an educated guess. And I believe that mechanical advantage(gears on the drive of the lift) have nothing to do with the amount of energy expended. If you have a low or high gear ratio, the same energy is used, just more quickly or longer.

I've been thinking all day and I find it hard to believe it onlt costs $12.60 a day to run the lift.. I think i goofed up.

Make it easier on yourself and just work in Kg. Going off your work tho, to get one chair to the top, PE=mgh=(170kg)(9.8)(311m), joules required to move one chair to the top, assuming its frictionless except for the drive mechanism, is 518126J 518126J / 360sec = 1439 watts per 2 riders, so about 700 watts per rider, x 2000 riders per day = 1400000watts in 7 hours (diablo last year was 9-4 right?) so 200000watts/hour, $.10kw/hr, and it comes out to $20,000 per day. Hmm. I think we're both wrong

Im going to say that at 32º the load is then halved, so 187.50lbs. F=ma=(187.50)(5.5)=1031.25ft.-lbs. So it takes 1031.25ft.-lbs. to move the cart up with 2 riders and 2 bikes. W=FD=(1031.25)(1992.65)=2054920.3ft-lbs.=.77391606kWh So it costs $.08/trip up the mountain. Using the average of 2000 trips/day it costs $154/day to run the lift or $16,170/season Also, just so you know the: potential kinetic energy is 485.814kgm gravitational potential energy is 528410.43kgm