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Kinetic Vehicle [flywheel powered] mass and energy

  1. Jun 27, 2017 #1
    1200lb total vehicle weight
    800lb total flywheel mass [4x200ea]
    Diameter 18.75"
    Flywheel RPM 3500 [120v electric motor plugged into wall]
    Pulley to drive wheel at 1:1 [6"pulley] or to 12' pulley

    Please help write the end of this story before I start. This is a "rev up" car built for big kids, in that it's plugged into the wall to spin up the flyweels [x4] to 3500 RPM. These RPM are transferred to the ground using kevlar belts and pulleys to the tire.
    I pull the lawnmower style belt clutch and dump the stored kinetic energy to the back tires directly onto the spines of my enemies. How many nemesis can I smear on a charge?
    Less useful answers might be -
    How would you rate the potential power to compare it to a 1/4 mile car? Newtons and spinning and procession mathubations aside, I want to know if the power is delivered to the tires how fast will I get in 1/4 or a mile based on the stored power and weight of the vehicle.
    I have done the calculations on every link in this forum, and used every calculator available that can translate from meters per second to dangles per minute. I realize aerodynamics and tire limitations are there, that's a real world penalty I don't need addressed either. What I have not been able to come to grips with is whether I'm working with enough stored energy or not. The force calculators all say that with the newtons available the vehicle should do over 3000mph. That's horsedirt. I'm off and totally wrong. Before I build framing I have to know how much of a bomb I'm driving.
    I have searched a month for the answer, even asked my NASA guy before growing the stones to ask here. Please think in the context of a flywheel toy that a grown up sits in, trying only to go super fast for 8 to 10 seconds or less.
    Forgive me if the answer is posted somewhere, I promise no answers I've seen help me so far. Thank you in advance for any help or suggestions you have.
     
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  3. Jun 27, 2017 #2

    CWatters

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    At some point you would have to take into account real world issues but....

    Top fuel dragsters have a known power output and known time for the quarter mile. So with simple maths you can make a crude estimate for the energy required. You could then compare that with the energy stored in your flywheel.

    You can also compare that to the energy produced by an explosive of your choice.
     
  4. Jun 27, 2017 #3

    jack action

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    The maximum speed your vehicle ##v_{v\ max}## can reach if all the flywheel energy is transferred is:
    $$v_{v\ max} = \sqrt{\frac{m_f}{m_v}} \omega_{f\ max} r_f$$
    For ##\omega_{f\ max}## = 366.5 rad/s (3500 rpm), ##r_f## = 0.238 m (9.375"), ##m_v## = 1200 lb and ##m_f## = 800 lb then ##v_{v\ max}## = 71 m/s (160 mph).

    But if we assume that your flywheel is constantly connected to your wheel with a given gear ratio ##GR## and wheel radius ##r_w## and that all the slipping occurs between the wheel and the road, then there will be a speed ##v_v## that will be reached. At that speed, there will be no slipping anymore. If there is no slipping, then further flywheel rpm reduction, will result in a wheel rpm reduction, thus a vehicle speed reduction as well. Therefore ##v_v## will be an actual maximum speed. The energy balance equation will be:
    $$E_{v @ v_v} = E_{f\ max} - E_{f @ v_v}$$
    $$\frac{1}{2}m_v v_v^2 = \frac{1}{2}m_f(\omega_{f\ max} r_f)^2 - \frac{1}{2}m_f \left( \omega_f r_f \right)^2$$
    Where ##\omega_f = GR \omega_w## and ##v_v = \omega_w r_w## such that:
    $$\frac{1}{2}m_v v_v^2 = \frac{1}{2}m_f(\omega_{f\ max} r_f)^2 - \frac{1}{2}m_f \left( GR\frac{r_f}{r_w}v_v \right)^2$$
    Rearranging:
    $$v_v = \sqrt{\frac{m_f}{m_v + \left( GR\frac{r_f}{r_w} \right)^2 m_f}}\omega_{f\ max} r_f$$
    Now, if ##\left( GR\frac{r_f}{r_w} \right)^2## if very small (close to zero), then ##v_v = v_{v\ max}##. If it is very large, then ##v_v## tends toward zero. Let's assume ##\frac{r_f}{r_w}## = 1 and that ##GR## could be either 0.1 or 10 as realistic extreme values. Then ##v_v## = 58 m/s (130 mph) or ##v_v## = 7 m/s (16 mph), respectively.

    At a low gear ratio (10:1), the wheel rpm is very slow, so the energy loss through slipping will be small. On the contrary, with a high gear ratio (1:10), the wheel rpm is very high, so the energy loss through slipping will be high. If we assume we have some continuously variable gear ratio from 10 to 0.1 when the vehicle reaches 16 mph, it would be possible to reach close to the 130 mph mark due to minimal energy loss through slipping (no slipping above 16 mph).

    If we assume a fixed gear ratio and that the best compromise between highest speed and minimum slipping is when ##GR## = 1 (right in the middle) and that only half of the ##E_{f\ max}## is available because the rest is loss through slipping (multiply ##m_f## of the numerator by 0.5), then ##v_v## = 32 m/s (72 mph).

    I cannot estimate the distance and time it will take to reach that speed without doing more complicated calculations and knowing more details about the vehicle, but the maximum speed is a good indicator for what you can expect.

    That would be my approach for a crude first approximation.
     
  5. Jun 28, 2017 #4
    Thank you for the formulas and your estimations, I very much appreciate it. You speak fluent math, I wish I could see as easily as you do. Just so I'm clear with you of all readers of this post - I am a player, not a musician. I mean that unlike the great postulators, I am willing to break things until it works instead of spending my time in advance deciding it won't. You might be the person who saves my fingers.
    I have attached a super rudimentary sketch here so you can see what I am physically planning. Feel free to pull the automatic shotgun out to pump this full of holes for me to plug, every one you see saves me a failed attempt, money, and depending on the RPM a lot more.
    There are 2 carriages, 1 on each side of the machine. These carriages each hold 2 flywheels and 2 electric motors. 1 motor spins the flywheels up, the other motor is also belt clutched to the front wheel. I dump the belt clutch from the flywheel to the rear tire [20" Hyabusa motorcycle tire] and it starts spinning the rears. In this design I've included a second electric motor that dumps the mechanical energy back into an electric motor on the front wheel for AWD..... maybe not the best setup.
    Would it make more sense to spin up the flywheels and dump the power back to the wheel through an electric motor instead of a belt and pulley or both?
    At 3500 RPM the outer rim velocity of the flywheels is far lower than mach, it shouldn't come apart at double that speed. The only real difference in what I can store is how long I feel like waiting before the iron starts coming apart.
    No, a 1/4 mile car is not an end goal even if it's the first prize. I am going to make a car that I can refuel anywhere I can plug an extension cord, or refuel using the air compressor at a gas station. Thank you for helping me smoke the Dodge Demons and ricers at the track, I'm starting the "no motor" class.
    Am I correct to assume I can build and release more power at once this way than the output of a combustion motor? If these flywheels doubled in velocity [quadruple stored power?] and then put to the ground I'm sure the numbers change too. I am justifying spending a lot of money soon with a builder, this is not for a class.

    Thanks for your insight.
     

    Attached Files:

  6. Jun 28, 2017 #5

    jack action

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    I thought the electric motors were not on the vehicle and were just used to «charge» your vehicle.
    This, I don't understand. The mechanical energy from where? An electric motor requires an electrical power source, how do you power it with mechanical energy?
    I'm no specialist in electrical power generation, but my first thought about creating a nice Continuously Variable Transmission (CVT) would be by powering a generator with your flywheels and feed electric motor(s) connected to your driven wheels.
    What is (are) your power source(s) exactly? Do you have a battery (or fuel tank or compressed air tank) on board as well?
    A flywheel is just like a battery and a fuel tank and it can hold a certain amount of energy. The rate at which that energy is used, that is power. Usually, power is set by the machine that transform the energy. For example, a combustion engine has the capacity for burning a predetermined amount of fuel per unit time. That also set a limit on how much power can be produced. But if you could just burn the entire fuel tank instantly, you would get an incredible amount of power (although for a very short period of time, since the tank would empty quickly). Same goes with the batteries of an electric car. The electric motor has the capacity to handle a certain current and voltage, which dictates the maximum power.

    As for flywheels, the power you can get is almost infinite because it is already mechanical energy and it doesn't need to be transformed like chemical or electrical energy. If you can stop them in a fraction of a second, the power output will be enormous. All you need are parts that won't break under such a load. But once stopped, the energy source is depleted.
     
  7. Jun 28, 2017 #6
     
  8. Jun 28, 2017 #7

    JBA

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    So then you probably destroy the tires without really going very far. Even the most powerful slingshot dragsters require power feed control at launch.
     
  9. Jun 28, 2017 #8
    Yes of course I want the same problems, I'm doing the same thing. If you can point me to a good mechanical idea to manage it, that would be great. The kevlar belts will slip until they grip, so at about $20 a belt I can afford destroying one after another. Any other ideas?
    If this eats tires and shoots down the pipe like a bat out of hell, this is good. We switch to metal wheels for the salt flats, in the meantime idiots on motorcycles keep a steady supply of fat rear tires in salvage yards when they exchange them for wheelchairs. [$120 for the entire rim/tire/brake setup].
    Thanks again for your help JBA, I do very much appreciate your contribution. I will no doubt include you on a list of people I have stepped on along the way, I always give the credit where it's due.
    But I'm keeping the money. Just so we're clear there. And the chicks too. None for you.
     
  10. Jun 28, 2017 #9

    JBA

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    Why not just keep it simple, forget the flywheel, put the flywheel weight on the wheels, lift them off of the ground, spin them up to speed and then drop them on the ground.
     
  11. Jun 28, 2017 #10
    Yeah, I had thought on that some. Know anyone that makes a 400lb tire?
    Seriously though, you can't spin up a tire to the same speeds you can an iron flywheel. The rubber will spin off, and Brembo doesn't make calipers to handle slowing down the intentions of a wheel that heavy spun up. Just over 220mph normal tires come apart, and the outer edge of these flywheels can handle 500mph or more before any problems. I don't think we can approach the same power levels putting a tire on a flywheel.
    By connecting them by belt and pulley, I can get an honest transmission of higher power loads without unsprung mass messing up my ability to steer or stop.
    Great idea though thanks. Any others?
     
  12. Jun 28, 2017 #11

    jack action

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    I don't know how you will be able to coordinate front & rear power/rpm outputs. You probably will start with the belt OR the electric motor and that will cause you enough problems to solve.

    You do realize that your installation is a very, very small energy reserve? The quantity of energy stored in your 800 lb flywheels @ 3500 rpm is 1.4 MJ. If you go on the highway and use only 14 hp (that is optimistic) to just cruise at legal speed, you have about enough energy to ride for 2 minutes.

    If you can maintain a traction force of 5300 N (1200 lb), i.e. equivalent to the weight of your vehicle, you have enough energy stored in your flywheels to travel about 260 m (850 ft). If you put drag slicks and fully use the maximum traction force they can handle, you can easily divide by 3 that distance.

    Start with that first. See how it works.
     
  13. Jun 28, 2017 #12
    You do realize that your installation is a very, very small energy reserve? The quantity of energy stored in your 800 lb flywheels @ 3500 rpm is 1.4 MJ. If you go on the highway and use only 14 hp (that is optimistic) to just cruise at legal speed, you have about enough energy to ride for 2 minutes.
    This understanding was slowly settling in.
    Using the calculator at this address [http://www.botlanta.org/converters/dale-calc/flywheel.html], I am given a bunch of numbers I can't translate as well as you just did. 12800oz [800lb] 18.75" diameter flywheel
    at 3500 rpm spits out these calculations - at 7000rpm these
    Disk KE (joules) 691407.8916036893 2765631.566414757
    Inertia (kg*m²) 10.29101916969147 10.29101916969147
    Ring KE (joules) 1382815.7832073786 5531263.132829514
    Inertia (kg*m²) 20.58203833938294 20.58203833938294
    Then there's the centrifugal force numbers etc
    Is there enough stored energy to cross 1320 feet? At miserable speed, or would the front wheels lift up?
    I realize at 7000rpm the speed is shy of 400mph surface speed on these discs, well within the flywheel's storage potential. I think there's another 100mph or so left of velocity before I have to upgrade to composites.

    It was just recently Newtons became more than just fruit and cake and not a cookie. Glad to meet you, I do very much appreciate your input. And that goes for everyone else, thank you.
     
  14. Jun 28, 2017 #13

    jack action

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    First, I just realized that I used the inertia of a ring and you are probably using a disc. So the time and distance of my previous post have to be divided by 2 if that is the case. And all the ##m_f## variable in my velocity equations must be multiplied by 0.5 as well.
    You can release the stored energy over 1000 miles if you wish; You just won't create a lot of power if you do so.

    The equations are the following:
    $$t = \frac{E}{P}$$
    $$F = \frac{E}{d}$$
    Where:
    • ##t## is the time it will take to use up the energy ##E## while spending power ##P## (in seconds)
    • ##E## is the energy stored in your flywheels (in Joules)
    • ##P## is the average power you spent during time ##t## (in Watts)
    • ##F## is the average traction force you used over traveled distance ##d## (in Newtons)
    • ##d## is the traveled distance by your vehicle (in meters)
    This is how I got the numbers from my previous post.
     
  15. Jun 29, 2017 #14
    You can release the stored energy over 1000 miles if you wish; You just won't create a lot of power if you do so.
    No, not 1k miles. 1320ft, 1/4 mile, roughly 402m.
    After looking at the links on your post, and thank you for more calculators - they postulate that what I want to do is possible, then bring up other problems that can be designed around [procession etc]. Instead of a 100hp motor like the example below, I'd use an extension cord to spin them up. I need the answer to his theoretical question though, in how big the flywheel should be and how fast you have to spin it up to get his bloated 3000lb car to a 10 second quarter. I wish this guy was more specific, but it doesn't seem anyone has any testing data for me to reference anywhere.

    Your guy at hpwizard says this:
    • The flywheel energy stored in first gear is "free". "Free" at the drag strip, anyway. Meaning time and energy spent spinning up the flywheel when staging doesn't count against your E.T. In fact, in the extreme case, a huge flywheel could actually make your car quite a bit faster. For instance, take a 3000lb car with a 100hp engine and an enormous flywheel. You could theoretically run a 10 second quarter mile...if you were allowed a few minutes to spin up the flywheel in the staging lane, and could find a way to dump the stored power without breaking anything. Like I said, this is an extreme example just to make a point. Flywheel energy storage systems can have tremendous power to weight ratios, but are a challenge for mobile applications due to gyroscopic forces and safety concerns.
     
  16. Jun 29, 2017 #15

    jack action

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    That is easy. The trap speed of a 10-sec 3000 lb (1360 kg) car is about 125 mph (55 m/s). The energy required to reach that speed from rest is:
    $$E = \frac{1}{2}mv^2$$
    $$E = \frac{1}{2} \times 1360 \times 55^2$$
    $$E = 2\ 057\ 000\ J$$
    Now you can go back to your calculator and find what combination of flywheel mass, radius and rpm will store at least that amount of energy. For a disc flywheel at 3500 rpm and a diameter of 18.75", you need a mass 2380 lb (which mean you will probably need to revise that 3000 lb weight for the entire car). If you double the diameter or the rpm, you can drop the mass to 595 lb.

    This doesn't take into account the energy losses and you still need to deliver that energy appropriately to get a proper wheel torque/rpm relationship that maximize the performance of the car.
     
  17. Jun 29, 2017 #16
    So then I can have confirmation from your side, if I can store 2million joules I can get a trap speed of 125mph from a 3000 lb car.
    My vehicle will weigh about 1200lb.
    At 7000 rpm, I'll be storing 2 765 631.5 joules in the flywheels [using the calculator referenced above].
    At less than 1/2 the vehicle mass, I should cross a 1/4 mile at a good clip. I don't know how fast, but you know for sure. I'd imagine 1/2 the mass should have an impact on the top speed and distance, no?
    I realize that the application of the flywheel momentum to the back tire is going to be tricky, but I plan on using a kevlar belt and pulley clutch system, like a riding lawnmower would use to engage its blades. I should be able to manage a take off by decreasing incrementally the belt slip until it's fully engaged, and just count on burning through $12 industrial strength kevlar v belts. Those aren't so easy to break, according to the literature. Just like in 1st gear, you can modulate the pressure until you get a full catch. At least I think. Burning clutches is a symptom of going fast.
    I'm going to redesign my carriages some based on the answers and recommendations you have given me. Thank you again for your help and attention, I'm already looking into newer composite flywheels that store ten times the power. They just cost ten times as much, I'd really like to avoid that if iron flywheels can get me there.
    From what I understand, accelerating eats much more energy than overcoming rolling resistance at a cruising speed, that it takes more to get to 45 mph than to stay there. I was thinking a cruising motor should be able to maintain speed in a similar way to a Prius. 10 electric hp is more than enough in that case to maintain speed at a level attitude. Extending the reach to this car will mean I have to unplug it, so I'll work on a street car later. I'm not sure how much power I'll need to overcome resistance at 70 mph in a 1200lb car, any ideas? Aero aside, of course.
     
    Last edited: Jun 29, 2017
  18. Jun 29, 2017 #17

    jack action

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    I' m going to correct my previous post as it takes slightly more energy than what I said. The energy I calculated is only for accelerating the vehicle, but it doesn't include the energy necessary to fight the aerodynamic force and rolling resistance.

    Once again, it is easy to estimate. A 1360 kg typical car that goes through a ¼-mile in 10.5 s requires an average power of about 400 whp (298 400 W; 1 Watt = 1 Joule per second). That kind of power applied during 10.5 s gives 298 400 X 10.5 = 3 133 200 J or about 50% more that I previously calculated. So the weight required for a disc flywheel with an 18.75" diameter, spinning at 7000 rpm, would be 906 lb to store that much energy.

    By the way @Nicodemus Rex , you don't need to contact HPWizard for this thread, I'm the guy behind the website.
     
  19. Jun 30, 2017 #18
    By the way @Nicodemus Rex , you don't need to contact HPWizard for this thread, I'm the guy behind the website.[/QUOTE]
    Awesome! Your site is in my fav's and I've used it a number of times. Don't fault my lack of connections, I will ask anyone I can for the answers I don't have. I'm more confident than before in your information. Being the guy who uses smart stuff is not the same as the guy who makes it, hats off.

    After playing some with the kid's fidget spinner, it's easy to feel where my problems are going to come up trying to lean these things when I steer. If I spin up the spinner and hold it horizontal, there is no resistance whatsoever to moving it right or left. I think if I spin these up vertically, the axles the flywheels are on are going to snap, shooting these flywheels into low earth orbit as I change directions. Do you think spinning them up horizontally will solve the procession issue?
    On that note, is it correct to assume a clockwise spun flywheel mounted horizontally would tend downward? Or is it the other way around?
    If I'm spinning up flywheels flat under the car, I would be able to afford much larger diameter flywheels, 48" instead. In 1960's Seattle they had a flywheel powered bus that used a huge one slung horizontal under the chassis that could charge for 3 min and go 10 miles.
    If I turn these horizontally to solve procession issues, I will no longer be able to belt clutch the power delivery vertically, this leaves me with using the mechanical energy stored in the flywheels to power electric motors.
    How strong an electric motor should I use then? I realize the fastest bike on the planet uses a 150 kw electric motor, those take about 550lbs to 60mph in about 2 seconds - could I power a couple of those up for a quarter mile?
     
  20. Jun 30, 2017 #19

    jack action

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    I'm not a big expert on gyroscopic effect, so I don't want to go too far in giving advice in that field. Have you read the Wikipedia page about FES? To solve the vehicle handling problem, I like the idea of 2 flywheels spinning in opposite direction.
    This is where I wish to help you for now. It is about understanding the different concepts of energy and power.

    A flywheel is like a fuel tank, it stores energy. A fuel tank doesn't have a power rating, an engine does. If you use a fuel tank of a given volume with an engine that produces, say, 50 hp, and it take an 1 hour to empty the fuel tank, then you know that if you use the same fuel tank with an engine producing 100 hp, it will last only 30 minutes. If you use it with an engine producing 5000 hp, it will take 36 seconds to empty it.

    An engine has the capacity to deliver a given amount of energy in a certain period of time. The faster it can deliver the energy, the more power it produces.

    Just like a fuel tank, you can use a flywheel with a motor that will produce any power rating you wish. The difference will be in how long it will take to «empty» your flywheel.

    So if you were designing a dragster with a combustion engine, you would first determine what is the force you can produce at the wheels. That is what the app on my website does. First, given the power rating at the wheel (no matter how that power is produced) and the speed of the vehicle, you can find out exactly what is the force at the tire at that particular speed. We also know that tire/road friction coefficient will determine a maximum force that can be applied, no matter the power rating. With this info (and some more) I can calculate the performance of the vehicle, like the time it takes to travel the ¼ mile distance.

    So once you found out the power rating that gives you the desired performance, you can choose the combustion engine that produces that power. This engine will have a given fuel consumption at that power rating, something like X gallons per hour. This will include energy lost in the transformation from chemical to mechanical energy (As you know engines are very inefficient that way). Since you know how much time it takes to do the ¼-mile, you can easily determine the number of gallons of fuel you need for the race. So you will choose the gas tank volume according to this number.

    It is the same for your flywheel: Once you have identified your needs power-wise and time-wise, once you have identified the losses while converting the energy going from the flywheel to the driven wheels, you will know how big (or fast) of a flywheel you need. The number I gave you for the energy requirement is what is needed at the wheels, so it doesn't take into account any losses between the flywheel and the wheels. The power I used to estimate the energy requirement was 400 hp (at the wheels) because that was what I estimated for being able to do a 10-second ¼-mile. (See how the power depends on what you want to do and not on how big is the fuel tank?)

    So if you use an electric generator/motor, it must be able to handle 400 hp (Don't forget that electric generator/motor are rated by the power they can produce continuously, not the maximum power they can produce). But an electric generator needs to spin at a certain rpm as well to produce its power; It must correspond with the flywheel rpm that is constantly slowing down. Probably with some sort of transmission with a (variable?) gear ratio.
     
  21. Jul 1, 2017 #20
    The great news is that the 150 kw motor is 200hp. Problem is that they weigh more than my aunt Liz. I don't need to add fat girls to the system to pass around the power from the flywheels to drive tires.
    I appreciate your gas tank vs motor analogy. The flywheels in this vehicle are only a gas tank until a belt clutch to the back tire is thrown. The rotational storage of energy is expressed and released as work, so the flywheels go from being a battery to a motor. The only release valves to this expression of power are the pulleys managing the transfer of the rpms to the tires.
    Your efforts are far from wasted. I understand that with a 1200 lb vehicle, 4 x 225 lb spun up to 7000 rpm is going to be more than enough power. At about 1500 or so rpm on the 20 inch motorcycle tire, I'll be doing over 100 mph. Somewhere between the 1500 rpm I want and the 7000 available with nearly 1/2 ton of intention behind it, the cage they sit in is going to fly. Stored in the rings is 6 222 671.024433203 joules or over 6 Mw. If I smear the 6 Mw entirely across a 1/4 mile in a car that weighs 1200 lb, it's going to be much quicker than a Tesla Model S 10.8 time. Or no? It's not going to take as much time to accelerate either, because an electric motor may have stronger torque than gas motors, but I understand torque on big flywheels is expressed as an absolute. I still wouldn't be able to translate what n/m value of torque is available, the numbers I come to are astronomical and can't be compared to anything in my frame of reference.
    I don't know that a CVT will work for the sake of accelerating and maintaining acceleration. It would have to accommodate from 18 in to maybe a 6 or a 3? An 18inch pulley to a 6 inch on the drive tire is cutting the rpm to 1/3, or 2300 rpm so - a belt run from the outer rim edge would give me about 160 mph worth of spin on the rear 6 inch or so no? I don't think I'd reach 80 because the flywheels would lose a ton of velocity, so I'd want to throw the 12 in or 18 in belt pulley as it slows down to increase rpm at the rear tire to 1/2 flywheel speed [when it's slowed to 1/2 at 3500] then equal to the flywheels as they slow. I think if I were to reach 100, it will take some time for power to bleed off even at a 1 to 1 exchange - speaking to the retardation of the flywheel rpm from rolling friction alone. Maybe a 3 in pulley and a 12 inch? This way if it cost me 4000 rpm to get to 100 mph, I could still throw the 12 in or 18 and keep going faster right?
    I'm starting to think 7000 rpm might be a bit much power now. Good thing I can test at 500 rpm.
    What's your take on how to mount these flywheels? I have been through about 6 months of wiki type articles, all written by people who certainly know their stuff. There seems to be absolutely NO test literature online I can find about someone who has mounted heavy flywheels to anything and driven around to see what happens. If I spin them in opposite directions, they should share an axle right? Wouldn't that bend it around as they fight on it? Flutter? If the bearing goes, or when the gear goes, the minute they knock off of each other in opposite directions, it would be a terrifying formula 1 wreck happening. Would the best arrangement be on the same axle but separated and spaced?
    To balance the vehicle properly, is it a good idea to have a pair of counterspinning flywheels at each corner then?
     
  22. Jul 1, 2017 #21

    jack action

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    I really need to insist on using the right terms in the right situation, because it is difficult to tell if you don't understand the concepts or you are just misusing the words:
    6 222 671.024433203 Joules is 6 million Joules or 6 MJ and it is a unit of energy.

    6 MW is 6 million Watts and it is a unit of power.

    1 Watt is equivalent to 1 Joule per second. If you deliver 6 MJ of energy within 1 second, then you produced an average of 6 MW (= 6 MJ / 1 s) of power during that 1 second. If you deliver 6 MJ of energy within 10 s (say, during a ¼-mile), then you produced an average of 600 kW (600 000 W = 6 000 000 J / 10 s) of power during that 10 s. Note that it is an average, so it is possible, for example, that you delivered 1000 kW the first 5 s and 200 kW the last 5 s, for an average of 600 kW [= (5*1000 + 5*200) / 10 ] for the entire 10 s. That would still mean you spent 6 MJ of energy during that 10 s. You can see that like you lift off your foot from the gas pedal after 5 s, thus using less power, thus using less fuel (i.e. energy).
    With the system of belt and pulleys you want to use, here how it will go. At rest, the driven wheels rpm is zero and the flywheel rpm is 7000 rpm. Once you engage the belt, something will necessarily slip. It will be either the belt on one pulley (or partly on both) or the tire on the ground or a combination of the belt and the tire slipping. In either case, the friction will eat up energy that will transformed into heat. This mean it will be lost and you won't be able to use that energy to propel the vehicle.

    Now, assuming the belt doesn't slip (meaning the static friction force of the belt/pulley system is stronger than the one from the tire/road contact patch), then the force that propels the vehicle is equal to the maximum friction force at the tire/road contact patch when the tire is slipping. Since the tire is fully slipping, it may never advance though.

    But, if you let the belt slip on the pulley, the friction force at the tire/road contact patch won't be enough to break the static friction of the tire and the tire will roll and propel the vehicle. The key is to keep that force below the breaking point to get the maximum force.

    The problem with that system is that until the rpm of the driven wheels is in synch with the rpm of the flywheel, there will always be slip and thus energy lost through heat that won't be used to propel the vehicle. Estimating that lost is not easy but it will be a lot. That is why you have a transmission with different gear ratios on a car: to synch the engine rpm with the wheel rpm as quick as possible. Otherwise, you would just play with the clutch, letting it slip as needed to have the needed force at the tire, which would not only waste fuel, but also destroy your clutch very quickly.
    Again, just to make sure we are all on the same page, rpm is a unit of angular velocity, not power.
    If their axles are parallel and their rpm are synch, it should be good (like if your flywheels were two gears connected together).
    If you only have one big flywheel and the bearing goes, you will have a terrifying wreck as well. Again, it is like a fuel tank. If you set fire to a 10 gal fuel tank or to two 5 gal fuel tanks, you will have the same amount of energy released. But if you have two 5 gal fuel tanks and only one is set on fire, then the fire will cause half the damage because only half the energy was released.
     
  23. Jul 1, 2017 #22
    Things get clearer and better every time I read a response, thanks again for your help.
    In humility I admit my "gription" on vocabulations is weak, and I may be hard of understanding on the things I have little experience with. As a relatively smart guy, I know that even retarded chinese children can speak chinese though, so of course it's easy to do. If I built a go cart in my garage that can make it to the convenience store a 1/4 mile up the way and back again without gas, then I have physically much more to show for my limited vocabulary than someone fluent or even an expert with a mouthful of beautiful ideas.
    The translations and numbers I am getting are from calculators using processes and ideas I have little to no experience with, I just have to trust that the numbers they give me as they are expressed reflect the right process I'm doing from my side. I get stuck for weeks in the right answers to the wrong questions. Thanks for your help with that.
    For example, when I said 7000 rpm is a lot of power, I meant that in the sense that there is nearly a 1/2 ton of weight spinning at that rate - the power stored in the flywheels comes out to likely be much more than I'll need for what I want to do. It's fantastic news, that was a deal killer for me.
    A professor did something similar to what I want to do in a Pinto, only mounting a more cylindrical flywheel in the rear seating compartment. You'll see from the article he and I have the same intentions, just different educations. You'd be amazed at what an idiot can accomplish, I surprise myself all the time. We are up to something similar, only I don't want to use a Pinto, gas motor, or geared transmission.
    http://www.autoline.tv/journal/?p=545
    345 kw is the total power output limit of the electric motors in the Model S tesla. It does a high 10 second 1/4. This means it's blowing that power down the track the whole time, 10 seconds of 345 kw is 3.45 Mw yes? The Tesla is bloated by heavy batteries too, it tips the the scales at over 4900 lbs.
    Even if I put carpet in the cockpit, I won't get near 1500lb in total weight and I'll have almost twice the power to spend over that time.
    I'll be sticking with 18.75" diameter flywheels mounted vertically, spun opposite and braced on shared axles. Of course they will be caged to contain any failures, and will be away from where the pilot sits in the center of the vehicle. I'm sticking with the idea of mounting the pilot between the 2 carriages that hold the flywheels, and make sure the cages holding the flywheels would fail in a way to contain them or eject them outward and away. Spectators are on their own.
    My challenge now that I know there is enough power to use, expressing it is going to be the next challenge. I need to rule out anything expensive, heavy, or non-mechanical. Since I haven't proven yet that I am limited to using a heavy gear reducer or a geared transmission I won't design it to use those things. I am thinking that in a 1/4 mile I will need 1 gear to get going, and another ratio to get fast. If I get unlimited torque, the big transfer should come in the second place, not first. There's going to be no avoiding overwhelming the rear tire.
    If I use the right pulley sizes from the flywheel to the rear tire, I should be able to predict the top speed on the given pulley based on the rpm the tire is delivering to the ground. Around 800 rpm on the rear tire is around 60mph or so. The next pulley should train the flywheels as much as possible, since I've probably lost a lot of energy.
    I wish I could predict how much power will come off of the flywheels to get to 60 mph. Then I could enter the power in reverse in these calculators to find out I would be left with 3500 rpm on the flywheels. The goal is to match the rear wheel rpm to the flywheel as soon as possible. If I engaged the pulley by the ratio that says [1 to 1 at 3500], it is aiming for 300mph with about 3Mw of power.
    Maybe it gets to 130. Maybe the guy in the Demon finished behind me. I don't know what ratios to use. I realize if I shoot first and aim later I'll find it, if you have any ideas on pulley ratios from the 18.75 to the 20 in tire I'd love to hear them. I have a basic understanding of how this works and zero experience.
     
  24. Jul 1, 2017 #23

    jack action

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    The right answer is 3.45 MJ but, yes. This is exactly what I did in post #17.

    As for the transmission, you do realize that the flywheel will constantly slow down as you wish that the wheels constantly speed up? How do you think you will achieve that without constantly changing the gear ratio?

    The Ford Pinto you talk about has a CVT connected to the flywheel. On that Wikipedia page you also have a link to the professor's contact page of his university.
     
  25. Jul 1, 2017 #24
    I'm thinking that in the transmission of power I will slip until the rpm are equal - either in the belt or between the tire and the road, then begin to slow at the rate of combined frictions once the rpm match up. I imagine first it will slip on the belt, then on the tire, then mad acceleration until the tire meets flywheel rpm delivered by the pulley. If I have reduced the rpm from the flywheel to 800 by the time the back time gets it with a 3 in pulley, then by the time I approach the top speed of that pulley's rpm - which should be near 50mph I can pull the 12 inch pulley and do it all over again from 3500 rpm on the flywheels [or whatever rpm is left in them].
    Is this thinking right?
    CVT's are expensive hardware and weigh more than kevlar belts. I only need 2 or 3 ratios total, I can figure those out if I knew in advance how many rpm melt from the flywheel to encourage the vehicle to the top speed of a 3 or 6 in pulley. I'd really like a low tech, low weight solution to transmission. If I figure it out, the kit would be easy for anyone to build.
     
  26. Jul 1, 2017 #25

    jack action

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    It can be done by having only 2 or 3 gear ratios, but your rpm will be synch only 2 or 3 time during the race (for a fraction of a second). This basically means that there will be slipping for the entire race. So you're burning rubber - or your clutch is slipping - through the entire ¼-mile. That doesn't seem to be a smart choice and you might need a lot more energy stored to compensate for this lost if you want to reach the end of the ¼-mile.
     
    Last edited: Jul 2, 2017
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