How do you measure the weight of a roller coaster car being pulled up the chain-driven lift hill? I know the weight of the loaded train is directly proportional to the power needed to pull the train up the lift hill so I suspect you can measure the current draw from the motor but beyond P=IV I am not sure what formulas exactly you would use to get the weight or mass (M) of the train from current (I). Any ideas?
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How do you calculate weight from current? 1
- Thread starter ntweisen
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You would need the power, not just the current and P is not always equal for VI - the phase angle must be considered, unless this is a dc motor. You would need a lot of other things as well, including friction and other losses. Measuring tension on the take up reel might be better.
David Castor
David Castor
Surely you can find physics equation for power required to lift a mass over a certain time. Make the lifting power equal to the electrical power, and you can find how mass and current are related.
If you want to get more accurate figure in the losses between electrical power consumed and mechanical powered delivered to lift the mass.
If you want to get more accurate figure in the losses between electrical power consumed and mechanical powered delivered to lift the mass.
- Thread starter
- #4
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- #5
So are you saying that you know that there MUST BE a formula for this, or are you saying that you already know the formula for this? If you already know it, then don't measure the current, get a proper transducer and measure the shaft power. They are readily available and not much more expensive than a current transducer.ntweisen said:
If you do not know the formula but know there must be one, then I suggest looking in some of the popular Engineer's Toolbox type websites that sell you such things, and look for a conveyor HP calculator. It will ask for the load and the angle of inclination, the two most important issues. It may ask for other things like belt width etc, which factors in friction, but for the most part whatever you use for a conveyor belt formula will account for a lot more friction than a roller coaster would, so it will estimate high in terms of the power requirement.
Here is one resource I know of that sells one.
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Or if this is for a real job, load up some cars with varying (but known) amounts of weight and measure the current - that way, you eliminate all the non-linearities, funny efficiency effects and odd corners in the characteristics of the transducers.
A
A
parvizsamadof
Mechanical
excuse me I don't think like you, I think you need the friction between chain and Sprocket wheels, the wheels and the rails and the critical point of your motion somewhere you need most of your power!
so on I think the weight is not enough!
I want, I can, I do.
so on I think the weight is not enough!
I want, I can, I do.
The torque of a shunt DC motor varies almost linearly with the armature current. Knowing the torque, gear reduction and diameter of the drive sprocket, you can calculate the force on the sprocket. This of course does not account for friction losses. But you should be able to get within 10% or so. If you need greater precision, you will need to go with load cells or something similar.
btrueblood
Mechanical
"so on I think the weight is not enough! "
Uh, your OP said (I paraphrase) "how do I measure the weight, knowing current"
Zeus answered (correctly) that you calibrate it using a known weight of the car and measuring current for that weight. Do it again for another, different known weight. You should then be able to interpolate between those two numbers to determine other, unknown weights, knowing current alone.
To do it any other way you must include all the effects, including the ones you listed, the ones Zeus listed, and the other effects that neither of you thought of...
Uh, your OP said (I paraphrase) "how do I measure the weight, knowing current"
Zeus answered (correctly) that you calibrate it using a known weight of the car and measuring current for that weight. Do it again for another, different known weight. You should then be able to interpolate between those two numbers to determine other, unknown weights, knowing current alone.
To do it any other way you must include all the effects, including the ones you listed, the ones Zeus listed, and the other effects that neither of you thought of...
LionelHutz
Electrical
We might be more help if we knew why the weight was needed.
I beleive many if not most large coasters are one-off designs to give the amusement park a unique ride.
If this is for something like setting a braking force further down the track then I would do tests as suggested once the coaster is built. You can start with an initial "best guess" and fine tune it using a few different weights during test runs.
If it is required before construction then you'll have to calculate the force required to pull the train up the hill and then calculate the corresponding torque at the motor shaft. Using the motor data, it should be fairly simple to calculate the current corresponding to the weight. Hopefully, you have some prior data from other designs to help pin a number on the expected losses.
I beleive many if not most large coasters are one-off designs to give the amusement park a unique ride.
If this is for something like setting a braking force further down the track then I would do tests as suggested once the coaster is built. You can start with an initial "best guess" and fine tune it using a few different weights during test runs.
If it is required before construction then you'll have to calculate the force required to pull the train up the hill and then calculate the corresponding torque at the motor shaft. Using the motor data, it should be fairly simple to calculate the current corresponding to the weight. Hopefully, you have some prior data from other designs to help pin a number on the expected losses.
- Thread starter
- #11
The reason for needing the weight is because there are actually two roller coasters on their own individual tracks which "duel" meaning there are several near-miss elements where the trains speed toward each other only to turn away at the last instance. One the trains are released from the lift hill (the highest point on the track) there is no way to control the speed. The idea is to weigh the trains and then use performance data to determine which train to release from the lift first and how much of a head start to give it.
I agree with the 'measure it' group. Weight the cars and run empty, then weigh some people (give them a free ride) and run them. Then sort out some big people weight them, load up the cars with them, and run.
Measure the current with each of these cases. This will give you the true weight verses current demand.
As btrueblood sez, just interpolate between these values and you will have a very close weight/amps correlation.
A side benefit will be the ability to track wear and adjustment shifts. During the morning dry runs keep a log of the empty cars currents. If they change dramatically or you see them trending it will be useful information.
Keith Cress
kcress -
Measure the current with each of these cases. This will give you the true weight verses current demand.
As btrueblood sez, just interpolate between these values and you will have a very close weight/amps correlation.
A side benefit will be the ability to track wear and adjustment shifts. During the morning dry runs keep a log of the empty cars currents. If they change dramatically or you see them trending it will be useful information.
Keith Cress
kcress -
LionelHutz
Electrical
OK, with that description I would do some weight testing once the coaster is built. I would imagine part of the qualification testing is putting weights in the seats and running the coaster.
You posted it is a DC motor so you likely already have a DC motor controller on it. That controller is also likely able to give you either an analog or communications output for percent load which you can use as feedback to the coaster controls.
You posted it is a DC motor so you likely already have a DC motor controller on it. That controller is also likely able to give you either an analog or communications output for percent load which you can use as feedback to the coaster controls.
parvizsamadof
Mechanical
I think its better you run a system in a dynamic model and then try to find the forces and after that you create the control system.
then run it in the real state by virtual masses and calibrate your system.
And I agree with BrianE22 in other wise its really dangerous!!
I want, I can, I do.
then run it in the real state by virtual masses and calibrate your system.
And I agree with BrianE22 in other wise its really dangerous!!
I want, I can, I do.
LionelHutz
Electrical
The OP posted there are near-miss elements implying the trains could crash but I hightly doubt the trains can actually crash. There will likely just be elements where the tracks come together like )( and the OP wants the trains to approach such elements at the same time for the greatest effect. If they miss the mark a little it will still be OK, just not as fun a ride as it could have been.
parvizsamadof
Mechanical
he can escape from crash for this works we have a lot of soft wares like ADAMS and ....
I want, I can, I do.
I want, I can, I do.
parvizsamadof
Mechanical
Dear Brian22 most of the time in mechanical we create the system in critical condition and then we calibrate system in the real condition that work.
I want, I can, I do.
I want, I can, I do.
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