Actually, this is not correct. It is very hard to imagine how "stacked springs" work, unless you have spent a lot of time designing electrical circuits. (truth!)
Here is how I would try to explain it:
Imagine two pillows stacked on top of each other. Two pillows are softer than one pillow.
Imagine that one is a feather pillow, and one is a foam pillow. When they are stacked together, they will feel softer than just one pillow.
Through a bunch of mathemagic, the formula ends up like this:
(feather pillow times foam pillow)
divided by
(feather pillow plus foam pillow)
equals
super soft pillow.
for resistors we have:
(300 ohms times 200 ohms)
divided by
(300 ohms plus 200 ohms)
equals
120 ohms
for stacked springs we have:
(300 pounds/inch times 200 pounds/inch)
divided by
(300 pounds/inch plus 200 pounds/inch)
equals 120 pounds per inch
This is really really hard to wrap your head around, so try to imagine the pillow scenario.
If you are very mathematical, then check this link:
here is a link for electrical resistance:
Here is a link for actual car springs:
Ask here if you have more questions. I can't promise I can answer right away, but I can help you understand how these things squish.
Jay (Ground Control)
PS. Just to really make things interesting: The same formula is used for volume of flow through two different sized pieces of tubing.
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Here is how I would try to explain it:
Imagine two pillows stacked on top of each other. Two pillows are softer than one pillow.
Imagine that one is a feather pillow, and one is a foam pillow. When they are stacked together, they will feel softer than just one pillow.
Through a bunch of mathemagic, the formula ends up like this:
(feather pillow times foam pillow)
divided by
(feather pillow plus foam pillow)
equals
super soft pillow.
for resistors we have:
(300 ohms times 200 ohms)
divided by
(300 ohms plus 200 ohms)
equals
120 ohms
for stacked springs we have:
(300 pounds/inch times 200 pounds/inch)
divided by
(300 pounds/inch plus 200 pounds/inch)
equals 120 pounds per inch
This is really really hard to wrap your head around, so try to imagine the pillow scenario.
If you are very mathematical, then check this link:
Adding a resistance in parallel to another resistance will
reduce the equivalent resistance.
Parallel Equation:
R1 = 10Ω
R2 = 20Ω
Req = R1R2/(R1+R2) = (1/R1+1/R2)-1
Req = 10•20/(10+20) = 6.67Ω
here is a link for electrical resistance:
Parallel Resistor Calculator R1 + R2 = R equivalent resistance circuit equivalent total resistor finder made easy piggyback = parallel - Eberhard Sengpiel sengpielaudio
Here is a link for actual car springs:
aFePower.com, aFe Power, aFe Filters, aFe, advanced FLOW engineering, Stage 2, Stage 2 Si, Stage 1, Super Stock System, Upgrades, Cold Air Intake Systems, Torque Tubes, Air Filters, Direct Fit, Direct Fit IRF, Intake Filters, Pro Guard D2 Fluid Filters, Fuel Filters, Oil Filters, Transmission Filters, Performance Exhaust Systems, CAT Back, Turbo Back, DPF Back, DPF-Delete, tips, Manifolds, Intake, Intercooler, Silver Bullet Throttle Body Spacers, Performance Packages, Pre-Filters, Cleaning Kits, aFe Gear, Mach Force XP, Large Bore HD, Pro 5R, Pro Guard D2, Pro Dry S, Takeda Intakes, Powersports, Aries AE1, AE2, Bladerunner, Magnum Force, Diesel Performance, BMW Performance, Bimmer Performance, Sport Compact Performance, Tuner Performance, Jeep Performance, NHRDA, T.O.P. Air Filter, Heavy Duty Air Filters, Tractor Filters, Diesel Truck Filters, Tractor Trailer Filters
Ask here if you have more questions. I can't promise I can answer right away, but I can help you understand how these things squish.
Jay (Ground Control)
PS. Just to really make things interesting: The same formula is used for volume of flow through two different sized pieces of tubing.
.
.
.
.
.
.
..
.
.
.
.


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