[johnsiddle]

Hi John,

I assume you mean a pump drive motor. Hydraulic motor typically refers to a device using hydraulic fluid flow to produce a mechanical output.

Contrary to what was told you on the other thread, main (lift) pump motors from fork trucks can be as powerful, or even more so, than traction motors. It depends on the type of lift truck. Just because the pump motor is rated on a low time-on duty cycle doesn't mean it won't run continuously, at a lower power output. And they are often fan cooled.

The pump motor from a forklift may be suitable for an EV conversion. Typically they are unidirectional, so be sure rotation direction is correct for you. Switching rotation requires internal modification. And typically the motor will have an internal splined shaft making a coupling for a car very difficult.

Regards,

major

Thanks Major.
The one I am looking at is on Ebay now, it is rated as continuous and has plenty of cooling from the picture it seems to have a fan and an open cover over the brushes, it still has the pump attached so presumably I could use parts of the pump to fabricate a flange of some sort, maybe weld a plate into it...
john

 

[johnsiddle]

I have picked up a Pump Drive motor from a forklift it isn't the one I was hoping to get but hopefully it will be suitable.
It has a female star shaft but I have the star bush that was fitted to a keyed taper shaft on the pump assy and I hope to fabricate a shaft with it.

Also it does not have a fan fitted but I am sure I could force some air in thro the brush guard, getting the hot air out again might be a problem tho.

It measures 11in by about 8in diameter and has a good clean commutator with four sets of twin brushes.
Rated at 67V and 115Amps but only at 15min.

There are three 8mm female connectors coming through the body which are marked A, YY and Y, I understand the A means armature ( I guess) but does anyone know what the YY and Y mean they have continuity to A connected internally. The guy I got them from said it was for forward and reverse.

Anyone know for sure what the connections mean and could it be a suitable EV motor.

John

 

[major]

Hi john,

From that info, I'd say it is a split field reversible series wound motor. The typical reversible series wound motor has 4 terminals, A1 & A2 for armature and S1 & S2 for the series field. Reversing is accomplished by reversing one or the other, but not both. This requires heavy (full motor current rated) contactors, two spdt. Read expensive.

The split series motor is actually designed and built with two independent (electrically) field coil sets. One coil set (Y) for one direction of rotation and the other coil set (YY) for opposite rotation. Now it can use a single spdt contactor or two spst. It does require additional diodes. This can be a cost savings. However you pay a penalty in series field resistive loss, perhaps affecting motor efficiency ~5%.

There are some other tricks designers can employ with the split series motor which can eliminate the reversing contactors altogether, do field weakening, and/or motor braking even with some degree of regeneration, when mated with a specialized controller.

Testing is the easiest method to verify what you have. Use a 12V car battery and jumper cables or the like. Put motor on floor or otherwise clamp it so it does not twist off the bench onto your toe. Connect one battery terminal to A. Connect (touch) the other battery terminal to Y. Note rotation direction. Leave A connected the same. Now touch other battery terminal cable to YY. If rotation is opposite, it is split series.

If rotation is the same, likely it is a tapped field. See if you can detect a difference in speed between the two connections.

Regards,

major

 

[J_D]

Hi Could somebody help me please?

Could someone tell me what voltage DC motor I would need to propel a 1400kg car to 70mph and the explanation behind it please too.


Thanks


Jon

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[johnsiddle]

Hi john,

From that info, I'd say it is a split field reversible series wound motor. The typical reversible series wound motor has 4 terminals, A1 & A2 for armature and S1 & S2 for the series field. Reversing is accomplished by reversing one or the other, but not both. This requires heavy (full motor current rated) contactors, two spdt. Read expensive.

The split series motor is actually designed and built with two independent (electrically) field coil sets. One coil set (Y) for one direction of rotation and the other coil set (YY) for opposite rotation. Now it can use a single spdt contactor or two spst. It does require additional diodes. This can be a cost savings. However you pay a penalty in series field resistive loss, perhaps affecting motor efficiency ~5%.

There are some other tricks designers can employ with the split series motor which can eliminate the reversing contactors altogether, do field weakening, and/or motor braking even with some degree of regeneration, when mated with a specialized controller.

Testing is the easiest method to verify what you have. Use a 12V car battery and jumper cables or the like. Put motor on floor or otherwise clamp it so it does not twist off the bench onto your toe. Connect one battery terminal to A. Connect (touch) the other battery terminal to Y. Note rotation direction. Leave A connected the same. Now touch other battery terminal cable to YY. If rotation is opposite, it is split series.

If rotation is the same, likely it is a tapped field. See if you can detect a difference in speed between the two connections.

Regards,

major

Thanks Major.
I have just checked and the main drive appears to be between A and Y. I can run between A and YY but it does go much slower PLUS I noticed that the connector for YY is only 6mm unlike the other two being 8mm.

Its strange it will run Y to YY albeit slow and A to YY still slow, but that might just be residual magnetism in the armature I guess in the Y/YY config.
It always runs the same direction.
Also it runs the correct direction for my gearbox (CW looking from the brushes end) so that is good.

I also put 50v on A/Y it and it sure took off, I am thinking when I get things running I will see how it runs using just 125V instead of the 174V I have available.

What are your thoughts generally about the suitability of this motor?

John

 

[kennybobby]

fissiks

Hi Could somebody help me please?

Could someone tell me what voltage DC motor I would need to propel a 1400kg car to 70mph and the explanation behind it please too.


Thanks


Jon

Sent from my F8331 using Tapatalk

How quickly do you need to accelerate up to the 70 mph, e.g. 0 to 70 in xx seconds? For a rough estimate this determines the power rating of your motor.

So first i'm gonna convert some units; 1400 kg is about 3100 lbs, divide this Weight by gravity, g, to get mass, m; and 70 mph is a velocity, v, of about 103 ft/sec.

Now the kinetic energy of motion at 70 mph is (.5mv^2) = 510682 lb-ft

For a first order power estimate take the kinetic energy divided by your acceleration time, let's assume 0 to 70 in 13 seconds.

This gives a power of 39283 lb-ft/sec, a horse can pull 550 pounds over a distance of one foot in one second, so divide by 550 to get about 72 horsepower required for acceleration.

Your specific question of motor DC voltage can not be answered without additional information, such as the size of your battery energy storage system available to supply power for your vehicle. The voltage and current rating of your battery pack is the deciding and/or limiting factor for the motor selection. But we can take a look at some variables involved.

72 hp multiplied by 746 gives you power in Watts, about 54000 W.

Now this power can be supplied as Volts times Amps = 54000.

If you have 1000 Amps available, then you only need 54 Volts, or
if you have 500 Amps available, then you only need 108 Volts, or
if you have 400 Amps available, then you only need 135 Volts, or ....etc.

See how this is going? there is no unique voltage solution until you select the limits for your acceleration time and your battery pack current. Then you can determine how many cells you need to make the voltage necessary, and then you can specify the motor winding voltage rating for the peak power that you need.

 

[J_D]

Wow, you must be a professor or something [emoji4], thanks so much for your post, it does actually make alot of sense in the way you've written it.

I'd like to propel the car to 60 in 6 seconds, with a top speed of 70mph.

I will have to get back to you about the batteries as I'll be using lead acid batteries as these are being provided free to me, until I can afford lithium.

I will provide the exact battery details and weights as soon as I am at work, hopefully you'll be able to tell me roughly If they need to be connected in parallel or series.

Thanks so much for your help I really appreciate it [emoji4]

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[kennybobby]

No, i'm just a ******* with a libary card...

 

[J_D]

[emoji1787]

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[Emyr]

Re: fissiks

So first i'm gonna convert some units; 1400 kg is about 3100 lbs, divide this Weight by gravity, g, to get mass, m; and 70 mph is a velocity, v, of about 103 ft/sec.

Now the kinetic energy of motion at 70 mph is (.5mv^2) = 510682 lb-ft

For a first order power estimate take the kinetic energy divided by your acceleration time, let's assume 0 to 70 in 13 seconds.

This gives a power of 39283 lb-ft/sec, a horse can pull 550 pounds over a distance of one foot in one second, so divide by 55

kinetic energy is measured in joules. lb-ft is a measure or torque, not energy.

Learn SI units, it makes the calculations a lot simpler.

 

[brian_]

units of measure

... Now the kinetic energy of motion at 70 mph is (.5mv^2) = 510682 lb-ft...

kinetic energy is measured in joules. lb-ft is a measure or torque, not energy.

In any system of units, the product of force and distance can be torque (moment of force), or energy, depending on context. For instance, in SI the units of torque are N⋅m, while the unit of energy is kg⋅m2/s2, or joule... which is also N⋅m. To keep these straight, the usual convention in the old units is to put the force unit first for torque (lb-ft), and the distance unit first for energy (ft-lb), so arguably kennybobby should have used ft-lb... but I don't think there's a consistent rule. Certainly in this context it is clear that the quantity is energy.

For a first order power estimate take the kinetic energy divided by your acceleration time, let's assume 0 to 70 in 13 seconds.

This gives a power of 39283 lb-ft/sec, a horse can pull 550 pounds over a distance of one foot in one second, so divide by 550 to get about 72 horsepower required for acceleration.

The fact that ft-lb is a unit of energy is demonstrated as kennybobby continued to show that the rate of change of energy is power... including that a "horsepower" is defined as 550 lb-ft/sec (just as a watt is a J/s or N⋅m/s or kg⋅m2/s3

Learn SI units, it makes the calculations a lot simpler.

Yes, that avoids the unit conversion constants, but the physics is the same regardless of the system of units.

 

[J_D]

Re: fissiks

How quickly do you need to accelerate up to the 70 mph, e.g. 0 to 70 in xx seconds? For a rough estimate this determines the power rating of your motor.

So first i'm gonna convert some units; 1400 kg is about 3100 lbs, divide this Weight by gravity, g, to get mass, m; and 70 mph is a velocity, v, of about 103 ft/sec.

Now the kinetic energy of motion at 70 mph is (.5mv^2) = 510682 lb-ft

For a first order power estimate take the kinetic energy divided by your acceleration time, let's assume 0 to 70 in 13 seconds.

This gives a power of 39283 lb-ft/sec, a horse can pull 550 pounds over a distance of one foot in one second, so divide by 550 to get about 72 horsepower required for acceleration.

Your specific question of motor DC voltage can not be answered without additional information, such as the size of your battery energy storage system available to supply power for your vehicle. The voltage and current rating of your battery pack is the deciding and/or limiting factor for the motor selection. But we can take a look at some variables involved.

72 hp multiplied by 746 gives you power in Watts, about 54000 W.

Now this power can be supplied as Volts times Amps = 54000.

If you have 1000 Amps available, then you only need 54 Volts, or
if you have 500 Amps available, then you only need 108 Volts, or
if you have 400 Amps available, then you only need 135 Volts, or ....etc.

See how this is going? there is no unique voltage solution until you select the limits for your acceleration time and your battery pack current. Then you can determine how many cells you need to make the voltage necessary, and then you can specify the motor winding voltage rating for the peak power that you need.

Hi Kenny

Right, I have 4x 12v 50ah battery's and 2x 12v 26ah battery's , would this suffice?

I may be able to get hold of some more, and what motor voltage would you recommend?

Thanks again

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[jbman]

Re: fissiks

Hi Kenny

Right, I have 4x 12v 50ah battery's and 2x 12v 26ah battery's , would this suffice?

I may be able to get hold of some more, and what motor voltage would you recommend?

Thanks again

Sent from my F8331 using Tapatalk

That's only going to get you 72 volts, and you are limited to the capacity of the 26ah batteries. Charging would be really awkward, too. You will need to match the capacity of your batteries. Those batteries would store about 3kwh, and you can only use a fraction of that as they are lead acid batteries.

You may want to reconsider your battery choice. 72v could potentially be the bare minimum, but most systems run at least twice that. It will not get you to your performance goal, especially when you factor in the weight of the batteries.

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[J_D]

Re: fissiks

That's only going to get you 72 volts, and you are limited to the capacity of the 26ah batteries. Charging would be really awkward, too. You will need to match the capacity of your batteries. Those batteries would store about 3kwh, and you can only use a fraction of that as they are lead acid batteries.

You may want to reconsider your battery choice. 72v could potentially be the bare minimum, but most systems run at least twice that. It will not get you to your performance goal, especially when you factor in the weight of the batteries.

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Thanks for your reply

So how about if I could double the amount of batteries I had, would that work? (ps could potentially not use the smaller 26ah batteries and upgrade to the 50ah instead)

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[kennybobby]

An Intuitive Approach to Fissiks and Mechanics

Take a look at your Craftsman torque wrench and see the units of torque: ft-lbs.

Every mechanic knows what a 5 lb sledge hammer feels like in the hand, but who has an intuitive feel for 15 Newtons?

So take a ratchet wrench that's 12 inches (1 ft) long and pull the handle with 5 or 10 or 50 lbs--that's gonna be 5,10 and 50 ft-lbs of torque.

How many mechanics have a 1 meter long ratchet in their toolbox? and what's the 15 N pull force feel like to generate 15 m-N of torque?

So here is my intuitive reason why the correct units for torque is ft-lbs or m-N.

Torque is a vector cross product defined as R x F = |R|*|F|*sin(theta).
Reference: https://en.wikipedia.org/wiki/Torque

The equation for Work (Work is equal to the change in kinetic energy), which is a vector dot product or scalar with the units of energy.
Work = F . L = |F|*|L|*cos(theta)
Reference: https://en.wikipedia.org/wiki/Work_(physics)#Mathematical_calculation
https://en.wikipedia.org/wiki/Work_(physics)#Work–energy_principle

Both torque and work appear to have similar units, but confusion can be avoided by following the order of the vector equations: ft-lb for torque and lb-ft for work or energy.

Or in metric units: m-N for torque, and N-m for work or energy (N-m is Joules in SI units).

 

[brian_]

Re: An Intuitive Approach to Fissiks and Mechanics

Every mechanic knows what a 5 lb sledge hammer feels like in the hand, but who has an intuitive feel for 15 Newtons?

15 newtons is the weight of 1.5 kg. That's far more familiar than "5 lb" for most people in the world outside of the U.S. Neither system is inherently easier; for each person, one system is more familiar.

How many mechanics have a 1 meter long ratchet in their toolbox?

Probably not many, unless they work on heavy equipment. On the other hand, doesn't everyone have a breaker bar about a metre long? I do.

 

[J_D]

So If I have 8x 12v 50ah , what would that give me?

Also What motor would be best suited to the battery capacity I have?

And would it help me achieve my 1400kg car goal of 70mph?

With a range of 30miles?

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[jbman]

Re: An Intuitive Approach to Fissiks and Mechanics

15 newtons is the weight of 1.5 kg. That's far more familiar than "5 lb" for most people in the world outside of the U.S. Neither system is inherently easier; for each person, one system is more familiar.


Probably not many, unless they work on heavy equipment. On the other hand, doesn't everyone have a breaker bar about a metre long? I do.

This conversation is probably better had elsewhere, as it's not very strongly related to finding a good motor.

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[jbman]

So If I have 8x 12v 50ah , what would that give me?

Also What motor would be best suited to the battery capacity I have?

And would it help me achieve my 1400kg car goal of 70mph?

With a range of 30miles?

Sent from my F8331 using Tapatalk

To find the capacity of your battery pack, multiply the voltage by the amp hours. In your case, those batteries in series would get you to 96 volts. Multiply that by 50 amp hours and you get 4800 watt hours, or 4.8 kilowatt hours. Assuming a moderate 350 watt hours per mile, considering your car is a little on the heavy side and you're adding a lot of lead on top of that, in an ideal scenario and draining your batteries completely (to 0, which would be very bad for them), you would only get 13.7 miles. I dont know if you'd make it to 70, but if you did you wouldn't be there long.

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[J_D]

Brilliant, thank you for your workings out, so in theory I would need more like 12x 50ah battery's to be safe.

Also in your opinion what, voltage DC motor would you choose? To try and get me to 60mph

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