@hummieee what if the whole vehicle weighs 1lb? 100lbs? 10,000lbs? 1,000,000lbs?
what if the ejected water ballast reaction mass is 10% of the vehicle weight? 50%? 75%? 99%?
what if the vehicle is traveling 500mph at the bottom of the first ramp? 750mph? 1000mph?
Well it’s super easy to answer all of those questions captain skidmark.
It doesn’t fucking matter.
@Brenternet Well lets assume the water weighs 20,000lbs and slope vehicle weighs 200lbs. I want to go 40 miles as fast as possible, without using any gasoline or electricity.
Let’s assume with the frontal area of the tank on the way down the ramp, its terminal velocity is 600mph.
If the 200lb slope vehicle was using regen braking the whole way down the 45 degree ramp to a depth of 2.5 miles, how many watt hours can they store in their battery from regen by the time they reach the bottom of the ramp while still maintaing 500+mph with the 20,000lb water tank?
Once they reach the bottom and release all the stored regen energy in one push with a linear motor against the 20,000lb 500mph tank, how fast is the 200lbs slope vehicle traveling? It certainly has enough kinetic energy to make it back up to the surface on the second ramp, and it wasn’t powered by any electricity or fossil fuels, just water.
How many watt hours would you have to use to go the same surface distance at the same velocity?
I’ll accept that answer. Obviously afterwards one would dump the water from the tank into a moistened hole at the bottom of the shaft to let it naturally evaporate and later retrieve the empty tank.
Yea I realized after I posted that, that if I were to take the lower kv motor and spin it faster with another motor that it would probably overheat and over volt. Worth a thought anyways 
Thanks
I think the next big innovation is going to be higher voltage systems.
But separate from that, moving away from the VESC platform. While it’s a great platform for DIYers and has built up a great ecosystem for derivatives, I think it’s being somewhat restrictive. I’ve learned a lot from my new job, and one of them is that most VESC’s and all derivatives are designed pretty poorly.
Oh, and on a side note, our motors suck. Like, all of them. I havent seen a single well-designed one for ESK8.
Decreasing airgap, beefier rear bearings, no circlips, high quality stators. We don’t have any of that, even the “custom” orders I’ve seen are still pretty lame. We should also be running on much higher pole pair counts.
Oh, and I want to see a mag encoder integrated into a motor. 3 hall sensors do not give you enough resolution.
Also, GT3 belts. It’s glorious. I’m running 15mm wide with 3mm pitch and I could not skip teeth even if I tried. Rolling resistance is miniscule.
Oh, and also spiral stators will decrease rolling resistance drastically. No more indexing or “cogging” when unpowered.
Just my 2 cents.
Conclusions
Using the same (202,000,000 J) amount of mechanical impulse energy at the top or bottom of the ramp, impulse at the top of the ramp resulted in 204,080,604.5 J KE at the bottom of the ramp while impulse at the bottom of the ramp resulted in 260,921,883.60 J KE at the bottom of the ramp, which is more than the total mechanical impulse energy.
Even after returning to the surface on a second ramp, the “impulse at bottom vehicle” had more kinetic energy than the total energy in the mechanical impulse, while the “impulse at top vehicle” had less.
The impulse at top vehicle did worse in terms of efficiency than a standard car pushing off the ground after returning to the surface, while the “impulse on bottom” vehicle did better.
Vertical Drop Length: 4123.69 meters (29sec @ 1g - 2.56 miles)
Free-fall Velocity at Bottom of Ramp: 284.39 meters per second (636.16 miles per hour)
Water Tank Mass: 10000 kilograms
Passenger Vehicle Mass: 100 kilograms
Mechanical Impulse Energy: 202,000,000 J
Impulse on Top (202,000,000 J):
m1 = 100kg
m2 = 10000kg
v1 = 2000m/s
v2 = 20m/s
m1v1 = m2v2
v2 = (m1v1)/m2
v2 = (100kg*2000m/s)/10000kg
v2 = 20m/s
m1 KE = (1/2)mv^2
m1 KE = (1/2)100kg*2000m/s^2
m1 KE = 200,000,000 J
m2 KE = (1/2)mv^2
m2 KE = (1/2)10000kg*20m/s^2
m2 KE = 2,000,000 J
m1 KE + m2 KE = 200,000,000 J + 2,000,000 J = 202,000,000 J = Mechanical Impulse Energy
Covering 4123.69 meters with initial 2000 meters per second velocity and 1 g acceleration takes 2.07 seconds.
2.07 seconds with 1 g acceleration is a velocity change of 20.3 meters per second
m1 velocity at ramp bottom with impulse on top: 2020.3m/s = 2000m/s + 20.3m/s
m1 KE (bottom of ramp, impulse on top) = (1/2)mv^2
m1 KE (bottom of ramp, impulse on top) = (1/2)100kg*2020.3m/s^2
m1 KE (bottom of ramp, impulse on top) = 204,080,604.5 J
Impulse on Bottom (202,000,000 J):
Free-fall Velocity at Bottom of Ramp: 284.39 meters per second (636.16 miles per hour)
m1 = 100kg
m2 = 10000kg
v1 = 2000m/s
v2 = 20m/s
m1v1 = m2v2
v1 = (m2v2)/m1
v1 = (10000kg*20m/s)/100kg
v1 = 2000m/s
V1’s actual velocity is 2000m/s + the 284.39 freefall velocity at the bottom of the ramp:
2000m/s + 284.39m/s = 2284.39m/s
m1 KE (bottom of ramp, impulse on bottom) = (1/2)mv^2
m1 KE (bottom of ramp, impulse on bottom) = (1/2)100kg*2284.39m/s^2
m1 KE (bottom of ramp, impulse on bottom) = 260,921,883.60 J
Conclusions:
Max Velocity with no ramp: 2000m/s (4473.87mph @ ground)
Energy with 202,000,000 J Impulse on top: 200,000,000 J <----
Max Velocity “Standard Car” pushing off ground: 2009.97m/s (4496.19mph @ ground)
Energy with 202,000,000 J Impulse on top: 202,000,000 J <----
Max Velocity with impulse on top of ramp : 2020.3m/s (4519.28mph @ bottom)
Energy with 202,000,000 J Impulse on top: 204,080,604.5 J
Ground Level Exit Velocity with impulse on top of ramp: 2000.18m/s (4474.28mph @ ground)
Exit Energy with 202,000,000 J impulse on top of ramp: 200,036,002 J <----
Max Velocity with impulse at bottom of ramp : 2284.39m/s (5110.03mph @ bottom)
Energy with 202,000,000 J Impulse on bottom: 260,921,883.60 J
Ground Level Exit Velocity with impulse at bottom of ramp: 2266.64m/s (5070.33mph @ ground)
Exit Energy with 202,000,000 J impulse on bottom of ramp: 256,882,844 J <----
^Even after returning to the surface on a second ramp, the “impulse at bottom vehicle” had more kinetic energy than the total energy in the mechanical impulse, while the “impulse at top vehicle” had less.
The impulse at top vehicle did worse in terms of efficiency than a standard car pushing off the ground after returning to the surface, while the “impulse on bottom” vehicle did better.
This was a nice flash back to physics class.
“oh hey, I remember that formula!”
Now go be a real engineer and fit all that math in the tiniest area possible with sloppy handwriting.
A bonus question on one of my exams ended up being something about a beer can on a block suspended by a spring. You had to calculate the minimum bullet weight shot by a gun at a certain angle and velocity to displace the can laterally and not vertically. Fun stuff.
I guess I can share this old idea 
(the gearing system don’t have to be huge (their s small efficient servo now), and if I have well listen, it possible directly from the Tx/Rx without vesc mod…)
You can read the whole discussion (all 5 pages of it) here:
Whew that Dale guy comes off a bit strong. Not a welcoming entrance to the physics forum haha. Reminds me a bit of a member here…
I enjoyed the read. A lot of valid points were brought up and I’m interested in your future findings.
Almost lost one of my DJI drones yesterday-- a mavic air, and quick formulation of a weird theory saved the drone…
I now know there is a hidden danger when flying a DJI drone in fog, even when you are in GPS mode.
Basically the drone has 2 vision sensor cameras front, 2 cameras rear and 2 pointing down which are only used internally for the drone’s orientation, landing and obstacle avoidance.
When flying in fog on default settings, as I discovered yesterday it is possible for the drone to “lose sight” of the ground. When this happens, and with landing obstacle avoidance turned on by default, the drone will simply start climbing. This will make you lose sight of the drone, leading to a situation where you can’t see the drone, nor can you see anything on the screen because of fog. A natural inclination might be to push down on the stick which lowers elevation, but when the drone can’t see the ground because of fog with landing obstacle avoidance on, it simply won’t descend.
So basically I was getting down below 20% battery, the telemetry said the drone was at 400ft, and 400ft horizontal distance, I couldn’t see it, I was pushing full down on the descend stick and the drone was not descending at all.
As panic began to set in at the thought of the drone running out of battery and simply dropping 400ft like a rock onto the street or worse someone’s house, car or heaven forbid a person or animal, I quickly formulated and executed a succesful plan.
I navigated deep into the menus of the drone’s flight settings and turned off all visual sensing, obstacle avoidance and landing obstacle avoidance, and then pushed full down on the stick for full speed descent…
… to my great relief the telemetry showed the drone rapidly descending, and I got it back on land with only a couple of % battery remaining. 
video:
What about the resources used to raise the vehicle to the top of the ramp and also pump the 20,000 lbs of water into the tanks. Nothing is free.
Here is what I envision:
The rider’s journey starts from ground level or slightly below ground level like entering a subway
The downward ramp is a tunnel 45 degrees, 2.5 miles depth since its achievable with existing technology, the deepest gold mines are 2.5 miles depth
The water is supplied from surface water resources, via gravity
The rider enters a lightweight capsule (for example 100kg total passenger vehicle mass including rider)
The aerodynamic tank is filled with water (for example 10000kg or ~2m edge cube equivalent)
The tank + passenger vehicle are dropped down the ramp
The tank has a “train” of small, lightweight roller cars with gear teeth in front of it and the passenger vehicle
If the ramp were vertical and a vacuum, the tank + passenger vehicle reach ~636mph after ~2.5 miles
There is a curved section of track at the bottom of the ramp leading to a horizontal section
Once the tank and vehicle are at 636mph at the bottom of the ramp, regen braking kicks in decelerating the tank + vehicle to 500mph, and the energy is stored in capacitors in the track and utilized at 70% efficiency, a process which can produce a usable 100 million joule mechanical impulse
The mechanical impulse is utilized to push the passenger vehicle off the 500mph 10000kg tank (not the ground), utilizing the “train” of gear teeth in front of the vehicle and co-moving with the tank
If the mechanical impulse were instead 202 million joules (19kWh) pushing off the 636mph tank (which is all I have done calculations for so far – suppose more water and a deeper tunnel were used), even after traveling back to the surface, the passenger vehicle has 256 million joules KE (~5070mph) which is more kinetic energy than the total mechanical impulse. It is also more kinetic energy than a standard car would have (202 million joules - ~4496mph) from pushing off the ground with the same mechanical impulse at ground level, a difference of nearly 600mph. This is known in astronautics as an “Oberth maneuver”
After the passenger vehicle pushes off the tank and returns to the surface via a second ramp, and is now traveling 5070mph on the surface, the tank no longer has sufficient kinetic energy to return all the way to the surface, so the water is instead dumped into a special chamber at the bottom of the tunnel, and the lightweight tank is retrieved with energy from harvesting more of the remaining kinetic energy in the tank at the bottom of the ramp
The nearly 10,000kg water now sits in a high surface area chamber at the bottom of the tank and heats up because the rock at that depth can be 150f from geothermal, and so it evaporates back into the atmosphere via a special air channel dug to the surface, to be later recycled
The 100kg passenger vehicle continues to its destination at more than 5000mph, having consumed no electricity or fossil fuels – fueled only by the gravitational potential energy of 10000kg water at ground level
^ @Skunk the simple version
I need the cliffsnotes for the entire conversation at this point. Idk what this is even a response to anymore.
I just saw meme opportunity
@professor_shartsis, you should provide an abstract(aka tldr) with the papers that you are submitting.
What about the consumption of resources to create your tunnel, how much energy is consumed to evaporate the water quickly enough to use this as a viable source of transportation for daily commute?
No free lunch.