TRA REO Grand SL SO LP
- Thread starter SeaNap
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TRA/SO/SL/LP
That's the only way I envisioned it....a HOT ROD
I'm so excited to get my prototype, It's been ready for a bit, but just had to wait for those contacts, so it should ship out this week as soon as I finalize my invoice...mini contacts, and other top secret items
sorry, sworn to secrecy
Seanap,
Your setup looks GREAT! Where did you get the Chaplin drip tip for your GOLD Cyclone?
Your setup looks GREAT! Where did you get the Chaplin drip tip for your GOLD Cyclone?
Thanks!
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It has a huge bore, it would be great for lung hits. I'm a mouth to lung kinda guy and with the AFC it is a bit much tbh
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...for now atleast. I can't wait for some company!Here are some numbers im getting;
Vbattery = 4.17V
Vbattery under load = 4.05V
Vcoil = 3.85V
VDtotal = 0.32V
Rcoil = .675Ω
Rbatt = 0.04Ω (from manufacturers spec sheet)
Rmod = 0.016Ω (WOW)
I = 5.703A (calculated)
Battery: Samsung INR18650 - 20R
Formulas used:
Vbattery = 4.17V
Vbattery under load = 4.05V
Vcoil = 3.85V
VDtotal = 0.32V
Rcoil = .675Ω
Rbatt = 0.04Ω (from manufacturers spec sheet)
Rmod = 0.016Ω (WOW)
I = 5.703A (calculated)
Battery: Samsung INR18650 - 20R
Formulas used:
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Check your paypal activity for tracking info. If it shipped Monday you might only need the spring, if it shipped yesterday it probably has everything. silver spring = old, gold spring = new, same with the top battery contact or what we call the firing pin.
Dang, SeaNap, I was trying SO hard not to order any more REOs
TRA SL grand (with upgraded contacts, of course) and silver button cover ordered *hangs head* I suppose I'll just have to wait a bit longer for the yacht I was planning to get
Besides, all those numbers you put in there fogged my brain and I can't be held responsible.
TRA SL grand (with upgraded contacts, of course) and silver button cover ordered *hangs head* I suppose I'll just have to wait a bit longer for the yacht I was planning to get
Besides, all those numbers you put in there fogged my brain and I can't be held responsible.
Thats just gorgeous that TRA SL LP SO! Thanks for the nice pics, its just a matter of time before I crack, probably get the mini 2.1 in black wrinkle, ahhh stop suggesting things to myself...!!! sshhhhh oh noooo, now ancient puffer's adding to the angst! 
Just a suggestion on the math, you can work out the internal resistance of your battery from the measurements you made:
R = V/I = (4.17 -4.05)/5.7 = 0.021Ω
i.e. when your battery is supplying 5.7A, the internal resistance is 0.021Ω.
The internal resistance of the mod, which is what the sub ohm kit is all about (reducing resistances within the mod) can be calculated with this:
Resistance of mod = {coil ohms x (BattV - AttyV)}/BattV
where BattV is the voltage at the battery terminals while firing the atty.
I get 0.033Ω, which is very close to mackman's 0.035Ω, post 1958 "mechanical reo's will be seeing some upgrades" thread.
The logic is this:
In a series circuit (which is what a mod has) the current through each component, battery, wire, switch and atty is always the same.
Since you calculated 5.7A through the atty, there is also 5.7A flowing through the battery, and through the small resistances of the mod.
Since the voltage at the battery terminals under load is 4.05V, and the voltage at the atty terminals is 3.85V, we have a 0.2V drop over the resistances within the mod.
We know there is 5.7A flowing, and a voltage drop of 0.2V, therefore 0.033Ω is the sum of resistive losses within the mod. The formula I stated above is just a re-arrangement of ohms law for this particular purpose.
If I've gone into too much detail here folks please forgive me, I can't help it
Just a suggestion on the math, you can work out the internal resistance of your battery from the measurements you made:
R = V/I = (4.17 -4.05)/5.7 = 0.021Ω
i.e. when your battery is supplying 5.7A, the internal resistance is 0.021Ω.
The internal resistance of the mod, which is what the sub ohm kit is all about (reducing resistances within the mod) can be calculated with this:
Resistance of mod = {coil ohms x (BattV - AttyV)}/BattV
where BattV is the voltage at the battery terminals while firing the atty.
I get 0.033Ω, which is very close to mackman's 0.035Ω, post 1958 "mechanical reo's will be seeing some upgrades" thread.
The logic is this:
In a series circuit (which is what a mod has) the current through each component, battery, wire, switch and atty is always the same.
Since you calculated 5.7A through the atty, there is also 5.7A flowing through the battery, and through the small resistances of the mod.
Since the voltage at the battery terminals under load is 4.05V, and the voltage at the atty terminals is 3.85V, we have a 0.2V drop over the resistances within the mod.
We know there is 5.7A flowing, and a voltage drop of 0.2V, therefore 0.033Ω is the sum of resistive losses within the mod. The formula I stated above is just a re-arrangement of ohms law for this particular purpose.
If I've gone into too much detail here folks please forgive me, I can't help it
TRA/SO/SL/LP
That's the only way I envisioned it....a HOT ROD
I'm so excited to get my prototype, It's been ready for a bit, but just had to wait for those contacts, so it should ship out this week as soon as I finalize my invoice...mini contacts, and other top secret items
TRA/SO/SL/LP? i just hear "trollop"-- as in "she's a beast"...
yes, my brain needs upgraded contacts
There is a mistake in my post #38 above, it should read
Resistance of mod = {coil ohms x (BattV - AttyV)}/AttyV
Resistance of mod = {coil ohms x (BattV - AttyV)}/AttyV
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