Power needed to move a vehicle against inertia

AI Thread Summary
The discussion focuses on determining the power and torque needed to move a man-powered front-wheel drive vehicle weighing 1,700 lbs, considering the effects of inertia and drag from the drivetrain. It is established that a minimum of one rider is necessary to break inertia, but achieving sustained movement requires more power, especially with additional riders. The vehicle's gearing ratio of 0.75:1 translates to a maximum torque of 1,200 lbs/ft from up to 12 riders, with peak power outputs varying based on rider cadence. Concerns are raised about the gearbox's durability under sustained torque, particularly during uphill movement or sudden braking, necessitating a robust design to withstand long-term use. Ultimately, the goal is to ensure the gearbox can handle the expected loads while maintaining a low vehicle profile and weight.
Electro Dave
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Hi,

I'm trying to figure out the minimum power in watts or HP to move a front wheel drive vehicle that is man-powered with pedals, of a given base weight and then to know how much people would be needed to pedal in order to break inertia and a constant of how much more power is needed for each additional person on board, considering a standard rider's weight and standard rider's power output.

The vehicle has 4 wheels, and only the front wheels are being driven. There is a constant drag from the whole gearing mechanism because the only freewheeling devices are at the pedals, which means the shafts and gearbox will always oppose resistance to the rotation of the front wheel. This will be added as an additional power rating on top of the basic power needed to break inertia.

The vehicle has a plain weight of 1 700lbs, including the driver (driver doesn't power the vehicle, only steers and brake);
Each additional rider weights 180lbs and can output roughly 100lbs/ft of torque up to a 40rpm cadence (0,76hp) and up to 90rpm (1,71hp) for a very short period of time, but still the max torque per rider is 100lbs/ft.
The gear ratio from the crankset to the wheel is about 0,75:1, meaning one full crank rotation will result in 3/4 rotation of the wheel. Friction coefficient is rubber on dry pavement.

A minimum of one rider has to be accounted to break inertia, but I highly doubt that 100lbs/ft of torque can be enough to break inertia of a 1 880lbs vehicle considering a 0,75:1 ratio (133lbs/ft to the wheels) ?

I'd like to know how much torque I would need at least to break inertia and how much power would be needed to maintain momentum, so I can more easily size the gearbox and calculate final gear ratio to reach a good top speed without needing too much torque from the riders to break inertia. We can have up to 12 riders, plus an additional 4 passengers, for a total of 16 plus driver, so a max weight of 4 580lbs and a max torque of 1 200lbs/ft from the riders.

On our current vehicle, we found that only 6 riders are needed to achieve a top speed, even when we are fully loaded at 4 500lbs, meaning they max out at about 120rpm at the crank, giving each a peak of 2hp for a short sprint period. Also, we found that 2 riders could break inertia, but couldn't maintain momentum for more than a very short period of time, because it required too much peak power.

My goal is to define the torque inflicted to the gearbox when breaking inertia and power needed to keep momentum.
 
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There is no minimum. In the absence of friction, any torque, no matter how small, will result in motion. It is just that for a small torque, the resulting acceleration will be small in direct proportion. As presented here, your question supposes that friction has already been accounted for elsewhere.
 
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Electro Dave said:
to break inertia.
There is nothing to break about inertia. You might be confusing inertia with static friction.
 
English isn't my first language, so I might not use the proper terms at time. I understand there would be a slight movement as soon as a force is applied, granted this movement could be considered non-existant when talking about the wheels starting to move the vehicle from it's initial position to a new position. I could describe everything better in French tho, but perhaps English is a more universal language on Physics Forums.

In short, let's say I use a 0,75:1 ratio from the cranks to the wheels, and each ride could give a peak torque of 100lbs/ft by standing up on the pedals when the wheels are locked, how many riders would it take to get the vehicle moving to a stable speed considering each rider can output 100lbs/ft to a max of 40rpm, and what would be that speed if 6 riders each give 100lbs/ft at 40rpm and the ratio to the wheels is 0,75 (133,33lbs/ft to the wheels per rider) ? Vehicle weight is 1 700lbs, each rider is 180lbs added weight.

I'm trying to eyeball a standard duty nominal requirement for my gearbox specs. If I use a gearbox that can handle up to 59hp at 2 400rpm with 1 550lbs/inch of torque on the high side and 3,7hp at 100rpm with 2 330lbs/inch of torque on the lower side of the rating table, will it be enough to withstand those 6 riders standing on the pedals that are linked to a same shaft going into the gearbox ?
 
The weight of the cart and of the riders is irrelevant. Ignoring air resistance and rolling resistance, there is no limit to the speed that even a single rider could obtain, given time.
 
Please, try to avoid the obvious and focus on the matter. As a cyclist, I know very well how much air resistance play into the equation and also that you can't move an object on Earth without a force strong enough to overcome rolling resistance. I can push the vehicle by hand fast easier than I can pedal it, but that's because I push in an almost linear direction whereas my own body weight on the pedals isn't enough to make it bulge.

Now, consider that we are in a big empty warehouse with no air movement (no wind) and a perfectly flat and levelled concrete surface that gives us minimum drag, tires are perfectly inflated and everything is well alignement for the shafts and bearings are of good quality and have minimal drag. The only restrictions are the gravity, the resistance of the air, the rolling resistance and the drag of the drivetrain. How much torque do my gearbox has to sustain before the wheels gain momentum ?
 
Electro Dave said:
Please, try to avoid the obvious and focus on the matter. As a cyclist, I know very well how much air resistance play into the equation and also that you can't move an object on Earth without a force strong enough to overcome rolling resistance. I can push the vehicle by hand fast easier than I can pedal it, but that's because I push in an almost linear direction whereas my own body weight on the pedals isn't enough to make it bulge.

Now, consider that we are in a big empty warehouse with no air movement (no wind) and a perfectly flat and levelled concrete surface that gives us minimum drag, tires are perfectly inflated and everything is well alignement for the shafts and bearings are of good quality and have minimal drag. The only restrictions are the gravity, the resistance of the air, the rolling resistance and the drag of the drivetrain. How much torque do my gearbox has to sustain before the wheels gain momentum ?
Any non-zero amount will do. If you want to be specific about how much rolling resistance there is, what cross section the device presents to the air or how fast you want it to go, that number could change.
 
We're not going to go anywhere at this pace.
 
Hmmh, how could I phrase it in a more obvious way... I want to know if this gearbox will be strong enough to last many years of party bike duty. It will be positioned on one side of the vehicle, collecting up to 6 riders on the cross shaft and transmitting to another central gearbox, as shown in the rendering below. Effective rpm range for this application will be between 0 and 120 at the crank. I consider the max torque and hp per rider to be of 100lbs/ft at any rpm from 0 to 40. There will be plenty of shock load as well.
Capture d’écran 2015-11-11 à 21.19.21.png

New_bike_1.png
 
  • #10
Please read post #2 again. jbrigs444 is correct. The power required depends on how fast you wish to accelerate.
 
  • #11
CWatters said:
Please read post #2 again. jbrigs444 is correct. The power required depends on how fast you wish to accelerate.

I do understand that concept. I should have said the torque that would be expected to make the whole thing move and then at which rate should I expect the bike to accelerate and the top speed I can expect to reach with a given max rpm and then how much continuous torque the gearbox will have to withstand when we are going uphill to a given steady rpm of the cranks.

I'm worried the gearbox will wear prematurely because of the sustained torque when we're in low rpm like going uphill or when the driver applies the brakes and people keep trying to pedal (drunk party people don't want to hear about braking, they want to pedal as fast as they can to reach the next bar haha). If I ask the owners to spend $3k on a gearbox setup instead of $1k on open spur gear and sprocket, I have to make sure it will last as long and longer than the spur gears and will be trouble-free and more quiet. But at the same time I'm trying to keep the floor of the vehicle as low as possible, meaning the bigger the gearbox, the higher the dance floor. And of course the weight, of the gearbox, the model 66 is already over 40lbs depending on the configuration, we want to keep the vehicle weight under a certain limit too.

Anyone can look at the gearbox specs and tell me if it will be fitted for the loads ?
 
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