layout power issue...i think

Add as many Lockons as needed, to account for the voltage drop due to the resistance of the track joints.  The wires feeding those Lockons should be as large as practical for the lowest voltage drop.

Your 75 watt transformer has about 50 watts output, the bulbs alone are using 16-20 watts of that leaving not much to run the train with, the engine with smoke will need about 20-25 watts, the whistle tender with its relay and motor another 20 watts. A freight set would not be as power hungry, lacking those 2 bulbs per car.

A set like this would ordinarily be used with a 1033 90 watts transformer, but you may want to look into a Type LW with an input rating of 125 watts and output voltage of up to 20 volts(compared to the 16 volts of the 1033 transformer).

Thank u for the reply Rob. That is exactly what  i needed to know.

Unless you test your power supply output voltage and current, you will never truly know the Watts(Power) it’s generating.
since Watts is a measurement of Joules transferred per second..and since joules are a measurement of energy under force, Voltage, and that force creates heat, when resistance is encountered…measurement of Watts, is not as easy as measuring Volts or Amps.
And since the majority of O Scale/Gauge trains are ALTERNATING CURRENT AC…your best bet is to focus on Current…which is measured as Amperes. If a 75 Watt power source is only provided 50 Watts-its time for a repair or a replacement😁

Do you have a Multimeter?

A decent digital meter is about $25 and worth having.
If you need help using it, to test your track, I’m available to instruct you.
I would upgrade your power source, then test your track.
A simple additional power drop maybe all you require

Good luck

@1drummer posted:

If a 75 Watt power source is only provided 50 Watts-its time for a repair or a replacement😁

75 watts is the input rating, 50 watts is the output rating.

@Roger439 it could be a couple of things...  Adding a lockon at the slowest point certainly can't hurt.  But you should also make sure that all of your track joints are tight, i.e., the pins fit snugly in their sockets on the end of the opposite rail, and there's no rust.  Also, is your platform level?  If the track is directly on the floor, is the floor level!?

Finally I would put a tiny drop of oil on each of your passenger car and tender axles.  Make sure the trucks swivel freely.  Friction is the enemy, and will definitely use up watts and cause a slowdown.  Good luck!

@1drummer posted:

Watts is a measurement of Joules transferred per second..and since joules are a measurement of energy under force, Voltage, and that force creates heat, when resistance is encountered…measurement of Watts, is not as easy as measuring Volts or Amps.

Uhh . . . it's *exactly* as easy as measuring volts and amps, since the wattage is just the product of voltage and amperage. In other words, if you are outputing 5 amps say, at 15 volts, that's 75 watts. Easy-peasy.

Thanks everyone for the great advice. I am a computer repair tech and have a general knowledge of electronics and electricity. I have had several ho layouts but never on this scale or this old so i really do aappreciate all the advice and help.

@Steve Tyler posted:

Uhh . . . it's *exactly* as easy as measuring volts and amps, since the wattage is just the product of voltage and amperage. In other words, if you are outputing 5 amps say, at 15 volts, that's 75 watts. Easy-peasy.

Hmm…yes, math is easy.
But, when I know my power source is providing 5 Amps

and I use my multimeter to measure the Amps

then the Wattage is redundant.

same thing applies to Voltage

Wattage is akin to foam in a beer mug

it looks nice

but that’s about it

especially in this situation where Alternating Current A/C is the premise.

@ADCX Rob posted:

75 watts is he input rating, 50 watts is the output rating.

I’m not familiar with that power supply.and it wasn’t designated by the author. Thanks for letting me know

@1drummer posted:

Hmm…yes, math is easy.
But, when I know my power source is providing 5 Amps

and I use my multimeter to measure the Amps

then the Wattage is redundant.

same thing applies to Voltage

Wattage is akin to foam in a beer mug

it looks nice

but that’s about it

especially in this situation where Alternating Current A/C is the premise.

Mmm, you have a very strange way of looking at things.

AC or DC, Ohm's Law still applies, as does the mathematical relationship between voltage, resistance and amperage. Wattage, whether expressed as a single number or as the product of voltage and amperage, is a measure of the energy being generated and consumed. Given two of the three, the final figure is fixed, and none are irrelevant to determining if the power supply is adequate for the task at hand. For instance, at track voltage of, say, 15 volts, five amps may be fine from a 75 or 100 watt-rated transformer, but try to draw five amps from a little 40 watt transformer and you'll either pop a circuit breaker or overheat/burn out the transformer coils. Beer foam, indeed . . .

@ADCX Rob posted:

75 watts is the input rating, 50 watts is the output rating.

@1drummer posted:

I’m not familiar with that power supply.and it wasn’t designated by the author. Thanks for letting me know

The various "Lionel Operating & Instruction Manual" issues printed over the postwar years each has a synopsis of the then available transformers.

The service manual documentation available on the LCCA website members-only area has detailed information on just about all of them. Wherever you can get access to this documentation, it is required reading for people interested Lionel transformer design, construction, input & output ratings, & repair.

In AC circuits, input or output power consumption can be considered in two ways.

Real Power = Watts (Average power over time)

Apparent Power = VA (Volt-Amperes).

Real power takes Power Factor into account, where Voltage and Current are out of phase.  In purely inductive circuits, AC voltage leads current by 90 degrees.  In purely capacitive circuits, the Current leads Voltage by 90 degrees.

For the sake of simplicity in model railroading applications, we often use Watts and VA interchangeably since they are somewhat close to each other and the power levels are relatively low.

Here's a link to more reading: What’s The Difference Between Watts And Volt-Amperes?

@SteveH posted:

In AC circuits, input or output power consumption can be considered in two ways.

Real Power = Watts (Average power over time)

Apparent Power = VA (Volt-Amperes).

Real power takes Power Factor into account, where Voltage and Current are out of phase.  In purely inductive circuits, AC voltage leads current by 90 degrees.  In purely capacitive circuits, the Current leads Voltage by 90 degrees.

For the sake of simplicity in model railroading applications, we often use Watts and VA interchangeably since they are somewhat close to each other and the power levels are relatively low.

Here's a link to more reading: What’s The Difference Between Watts And Volt-Amperes?

Thanks

But it hasn’t changed since I studied it to get my degree….so thanks

One other possibility not yet mentioned: if one of the cars creates a short, the whole train won't run regardless of size of transformer or number of power feeds.