1LikesElectric 1/8th Scale conversion kits.
11-29-2008, 04:17 AM
I personally like the standard tray better, it is lighter and you can movet it closer to the center of the car.
11-29-2008, 04:31 AM
Unfortunately, the Mugen is about the only one the plastic will not fit. The out-drive shaft on the Mugen is a smaller diameter than every other manufactures. Your ok with the metal to metal.
I just put back my plastics in place of the metal to metal for the indoor track. When all others have plastic, it sounds like mine is ready to blowup but it is really just fine.
11-29-2008, 07:21 AM
Tech Adept
Nue bearings
I need some help guys, I have a NEU1512 2d witha bearing going out. where do you get new bearings? I cant find them. Thanks
11-29-2008, 07:29 AM
I have a great deal on a 8ight b + rcpd roller with extras.
http://www.rctech.net/forum/r-c-item...ml#post5105900
Just an FYI,
http://www.rctech.net/forum/r-c-item...ml#post5105900
Just an FYI,
11-29-2008, 07:36 AM
Ohms law;
P(power)=U (voltage) x I (current)
Total wattage pulled by the motor = (Voltage) times (amp draw)
Without regard to other contributing factors, increasing the voltage will result in lower amperage draw for the same total output of power.
Lower amp draw (i.e less current flowing through the system) means there will be less power lost to heat from the resistance of the system. Starting with the IR of the battery, the resistance of the wiring from batt to ESC, the resistance of the ESC itself, and wiring running from ESC to motor. The higher the voltage, the less the internal resistance becomes a factor.
11-29-2008, 08:20 AM
11-29-2008, 08:37 AM
No issue really. It's a matter of efficiency and reducing "wasted energy". Basic physics
Ohms law;
P(power)=U (voltage) x I (current)
Total wattage pulled by the motor = (Voltage) times (amp draw)
Without regard to other contributing factors, increasing the voltage will result in lower amperage draw for the same total output of power.
Lower amp draw (i.e less current flowing through the system) means there will be less power lost to heat from the resistance of the system. Starting with the IR of the battery, the resistance of the wiring from batt to ESC, the resistance of the ESC itself, and wiring running from ESC to motor. The higher the voltage, the less the internal resistance becomes a factor.
Ohms law;
P(power)=U (voltage) x I (current)
Total wattage pulled by the motor = (Voltage) times (amp draw)
Without regard to other contributing factors, increasing the voltage will result in lower amperage draw for the same total output of power.
Lower amp draw (i.e less current flowing through the system) means there will be less power lost to heat from the resistance of the system. Starting with the IR of the battery, the resistance of the wiring from batt to ESC, the resistance of the ESC itself, and wiring running from ESC to motor. The higher the voltage, the less the internal resistance becomes a factor.
Actually, if I am reading what you are saying correctly, this is not true. Heat does not come from amps alone. Heat is a function of power which is a function of voltage AND current. power is energy. 4V * 5amps is the same power as 5V * 4amps. The two of these combinations will generate the same amount of heat. Same power, same energy. it's the power that matters, not just the current nor not just the voltage. it's the combo of the two.
reducing the current and increasing the voltage is the same thing as increasing the voltage and reducing the current. keep in mind here, when we are reducing the current of the of a lower KV motor, we are increasing the voltage to say, 4S to 5S. same power as a higher KV motor with higher AMPS with a lower 4S battery.
http://en.wikipedia.org/wiki/Electrical_power
Quotes from wikipedia
Electric power is defined as the rate at which electrical energy is transferred by an electric circuit. Devices convert electrical energy (hint power) into many useful forms, such as heat (electric heaters), light (light bulbs), motion (electric motors), sound (loudspeaker) or chemical changes.
UPDATE
but with this being said, there is a reason why the lower KV motors are more efficient. Here is an example:
NEU 1515 2D w/ 4S running @ 30340 RPM
vs.
NEU 1515 2.5D w/ 5S running @ 30525 RPM
2D is rated at 125A. Power = V*I = 4S * 3.7 * 125A = 1850watts
2.5D is rated at 78A. Power = V*I = 5S * 3.7 * 78A = 1443watts
Diff = 407watts
since the amperage is so much lower between the 2D and 2.5D vs the voltage going from 4s to 5s, you save a lot more power for about the same RPM.
Last edited by teeforb; 11-29-2008 at 09:05 AM.
11-29-2008, 09:29 AM
It's killing me but I'm going to miss out racing tonight. They are running for sure? Sounded pretty iffy. If you go - good luck!
And yes you are correct the resulting power is determined by both current and voltage. Just as the equation states. My point was that efficiency is lost through the system as a result of electrical resistance and internal resistance of the battery. Resistance becomes a larger factor as the current increases.
More wiki quotes
Electrical resistance is a ratio of the degree that an object opposes an electric current through it.
Internal resistance is a concept that helps model the electrical consequences of the complex chemical reactions inside a battery. When a current is flowing through a cell, the measured e.m.f. is lower than when there is no current delivered by the cell. The reason for this is that part of the available energy of the cell is used up to drive charges through the cell. This wasted energy is the so-called "internal resistance" and shows up as lost voltage.
And yes you are correct the resulting power is determined by both current and voltage. Just as the equation states. My point was that efficiency is lost through the system as a result of electrical resistance and internal resistance of the battery. Resistance becomes a larger factor as the current increases.
More wiki quotes
Electrical resistance is a ratio of the degree that an object opposes an electric current through it.
Internal resistance is a concept that helps model the electrical consequences of the complex chemical reactions inside a battery. When a current is flowing through a cell, the measured e.m.f. is lower than when there is no current delivered by the cell. The reason for this is that part of the available energy of the cell is used up to drive charges through the cell. This wasted energy is the so-called "internal resistance" and shows up as lost voltage.
11-29-2008, 09:32 AM
i just talked to Dan. Racing is on!!!
Hopefully u can make it for the toys for tots
Hopefully u can make it for the toys for tots
11-29-2008, 09:56 AM
I^2 * R
125^2 * .006(motor resistance) ~93 watts lost
78^2 * .008 (motor resistance) ~48 watts lost
125^2 * .006(motor resistance) ~93 watts lost
78^2 * .008 (motor resistance) ~48 watts lost
Last edited by az00lude; 11-29-2008 at 10:14 AM.
11-29-2008, 10:15 AM
Hey rice_man, hope to see you at the track today!!!
Actually, if I am reading what you are saying correctly, this is not true. Heat does not come from amps alone. Heat is a function of power which is a function of voltage AND current. power is energy. 4V * 5amps is the same power as 5V * 4amps. The two of these combinations will generate the same amount of heat. Same power, same energy. it's the power that matters, not just the current nor not just the voltage. it's the combo of the two.
reducing the current and increasing the voltage is the same thing as increasing the voltage and reducing the current. keep in mind here, when we are reducing the current of the of a lower KV motor, we are increasing the voltage to say, 4S to 5S. same power as a higher KV motor with higher AMPS with a lower 4S battery.
http://en.wikipedia.org/wiki/Electrical_power
Quotes from wikipedia
Electric power is defined as the rate at which electrical energy is transferred by an electric circuit. Devices convert electrical energy (hint power) into many useful forms, such as heat (electric heaters), light (light bulbs), motion (electric motors), sound (loudspeaker) or chemical changes.
UPDATE
but with this being said, there is a reason why the lower KV motors are more efficient. Here is an example:
NEU 1515 2D w/ 4S running @ 30340 RPM
vs.
NEU 1515 2.5D w/ 5S running @ 30525 RPM
2D is rated at 125A. Power = V*I = 4S * 3.7 * 125A = 1850watts
2.5D is rated at 78A. Power = V*I = 5S * 3.7 * 78A = 1443watts
Diff = 407watts
since the amperage is so much lower between the 2D and 2.5D vs the voltage going from 4s to 5s, you save a lot more power for about the same RPM.
Actually, if I am reading what you are saying correctly, this is not true. Heat does not come from amps alone. Heat is a function of power which is a function of voltage AND current. power is energy. 4V * 5amps is the same power as 5V * 4amps. The two of these combinations will generate the same amount of heat. Same power, same energy. it's the power that matters, not just the current nor not just the voltage. it's the combo of the two.
reducing the current and increasing the voltage is the same thing as increasing the voltage and reducing the current. keep in mind here, when we are reducing the current of the of a lower KV motor, we are increasing the voltage to say, 4S to 5S. same power as a higher KV motor with higher AMPS with a lower 4S battery.
http://en.wikipedia.org/wiki/Electrical_power
Quotes from wikipedia
Electric power is defined as the rate at which electrical energy is transferred by an electric circuit. Devices convert electrical energy (hint power) into many useful forms, such as heat (electric heaters), light (light bulbs), motion (electric motors), sound (loudspeaker) or chemical changes.
UPDATE
but with this being said, there is a reason why the lower KV motors are more efficient. Here is an example:
NEU 1515 2D w/ 4S running @ 30340 RPM
vs.
NEU 1515 2.5D w/ 5S running @ 30525 RPM
2D is rated at 125A. Power = V*I = 4S * 3.7 * 125A = 1850watts
2.5D is rated at 78A. Power = V*I = 5S * 3.7 * 78A = 1443watts
Diff = 407watts
since the amperage is so much lower between the 2D and 2.5D vs the voltage going from 4s to 5s, you save a lot more power for about the same RPM.
1515 3D rated @ 60A and 30192 RPM. Power = V*I = 6S * 3.7 * 60A = 1332watts.
Even more efficient compared to the 5S setup.
and to correlate this data with runtime, lets say 6000mah is max between 4S, 5S, and 6S. since you will use less wattage with a 6S system, you will have great run times comparing 4S and 5S system.
11-29-2008, 10:23 AM
Power transfer.... think of HV transmission lines.
They use high voltage so that the power will transfer efficiently. I think engineers have done their home work. Resistance use up a whole lot of power when trying to transfer power. Higher voltage is just better.
They use high voltage so that the power will transfer efficiently. I think engineers have done their home work. Resistance use up a whole lot of power when trying to transfer power. Higher voltage is just better.
11-29-2008, 10:25 AM
the whole 4s vs 5 or 6 s thing is something to think about. i am not a rocket scientest and dont know all the details. but if you have a 4s 6000mah system with say a 3000kv motor setup and have found the right gearing for it. then compare it to a 6 s 4000 pack with a 2000 kv motor from the same company with the same gearing, the 6s system will perform basicly the same on the track but will run cooler and will give a little better runtimes.
this has basicly been proven by the touring car guys when they went from 6 cell to 5 cell nimh overseas. in order to get the cars back up to speed they had to gear it higher and they started to pull serious amps and started to cook everything, getting hot enough to desolder the motor leads.
i personally run 4s and it seems fine but i still believe that 6s with the correct lower kv motor would end up being the best setup for 1/8 buggy. unfortunatly if roar holds to its 4s rule very few people will continue to develop this option and 4s will end up the norm.
if you think i am wrong about this then go and get a motor just like the one you run now but with double the kv and get a battery just like the one you run now but with twice the mah and only 2s. in theory the setups should be the same but i bet you will put the hurt on the battery and motor.
this has basicly been proven by the touring car guys when they went from 6 cell to 5 cell nimh overseas. in order to get the cars back up to speed they had to gear it higher and they started to pull serious amps and started to cook everything, getting hot enough to desolder the motor leads.
i personally run 4s and it seems fine but i still believe that 6s with the correct lower kv motor would end up being the best setup for 1/8 buggy. unfortunatly if roar holds to its 4s rule very few people will continue to develop this option and 4s will end up the norm.
if you think i am wrong about this then go and get a motor just like the one you run now but with double the kv and get a battery just like the one you run now but with twice the mah and only 2s. in theory the setups should be the same but i bet you will put the hurt on the battery and motor.
11-29-2008, 10:37 AM
Here is a good explanation
http://www.madsci.org/posts/archives...8929.Ph.r.html
Why is high voltage more efficient at transferring electricity?
Ben,
Thanks for the question. The formula that gets quoted to you is probably I^2 R
= P(line). The context for this equation is a part of an electrical circuit,
like a transmission line, which has a current denoted I flowing through it. The
wire in the line has a resistance R. What the equation is saying is that the
power lost to the resistance of the wire is I multiplied by itself (squared) and
then multiplied by R.
This equation has some implications. Obviously, if the resistance is big, then
you will lose a large amount of power to the line; it heats up. The I^2 part
means that having a large quantity of current flow through the line is really
bad for power loss, doubling the current makes the power loss 4 times as big.
To minimize power loss you want the current to be as low as practical.
Now on to your question. The power that is delivered by a transmission line is
described by the equation VI = P(delivered). Here V is the voltage of the line
and I is the current flowing throuhgh the line and into whatever is using the
power delivered. For a fixed amount of power that is needed, V and I can take
on many pairs of values, just as long as the their product VI is the power
needed.
Remember that to keep the power loss in the line low it is best to make the
current I small. When we do this it means that the voltage V has to be large so
that the delivered power is the needed amount.
We don't have to use high voltages to transmit electrical power. We could go
with low voltages and high currents. However, this would mean that the power
loss in the lines would be large unless the resistance was small. This would
then mean that we would have to use very thick and expensive wires to carry
electricity, so that the lines wouldn't heat up and melt.
I hope that this answers your question about high voltage electrical power
transmission.
Regards,
Everett Rubel
Current
http://www.madsci.org/posts/archives...8929.Ph.r.html
Why is high voltage more efficient at transferring electricity?
Ben,
Thanks for the question. The formula that gets quoted to you is probably I^2 R
= P(line). The context for this equation is a part of an electrical circuit,
like a transmission line, which has a current denoted I flowing through it. The
wire in the line has a resistance R. What the equation is saying is that the
power lost to the resistance of the wire is I multiplied by itself (squared) and
then multiplied by R.
This equation has some implications. Obviously, if the resistance is big, then
you will lose a large amount of power to the line; it heats up. The I^2 part
means that having a large quantity of current flow through the line is really
bad for power loss, doubling the current makes the power loss 4 times as big.
To minimize power loss you want the current to be as low as practical.
Now on to your question. The power that is delivered by a transmission line is
described by the equation VI = P(delivered). Here V is the voltage of the line
and I is the current flowing throuhgh the line and into whatever is using the
power delivered. For a fixed amount of power that is needed, V and I can take
on many pairs of values, just as long as the their product VI is the power
needed.
Remember that to keep the power loss in the line low it is best to make the
current I small. When we do this it means that the voltage V has to be large so
that the delivered power is the needed amount.
We don't have to use high voltages to transmit electrical power. We could go
with low voltages and high currents. However, this would mean that the power
loss in the lines would be large unless the resistance was small. This would
then mean that we would have to use very thick and expensive wires to carry
electricity, so that the lines wouldn't heat up and melt.
I hope that this answers your question about high voltage electrical power
transmission.
Regards,
Everett Rubel
Current
11-29-2008, 11:19 AM
I use a Neu 1515/2Y, my own 10S 120A car ESC and 8S (2 4S 3200-3700mAh) lipos in my MBX5T and have only seen spikes of 60A. If no one has tried HV, please do so with the right motor, not a 2200Kv with 6S. My other set up is a BPP truggy with a Neu 1521/1Y which I use 5S (2 TP 3850mAh 5S or 2 TP 4600mAh 5S in parallel), 6S (2 Kong Power 5000mAh 3S packs in series) and my Schulze 40.160. I've seen anywhere from 80-100A spikes with that set up. I've heard of guys going as high as 180A+ spikes with their 4S set up.