There is no way to predict accurately exhaust geometry without a good computational engine simulation. Too much variation mainly due to the exhaust gas temperature varying from one end to the other. What you are trying to do is to size the area to achieve a certain flow velocity and a length to create a path for the pressure waves. These pressure waves travel at the speed of sound ... and speed of sound depends on the gas temperature ... which varies across the pipe length! So the more you simplify, the more assumptions have to be made. These rules of thumb are as good as you're gonna get. Le me explain to you where they come from, such that you understand them better. Basics When the valve opens, a huge pressure wave is sent down the exhaust pipe. Pressure waves are the way gas molecules «communicate» between each other. The message that is sent is «We have lot us coming thru, make some space». This message travels at the speed of sound in all directions. The higher the pressure inside the cylinder, the stronger the pressure wave. Once it reaches an area change, the pressure wave continues to go on, but at a weaker pressure. The larger the area change, the weaker the signal. But it also sends back another pressure wave, but this time a really weak one (think vacuum). This one goes back to the valve with the following message: «We found some free space, it's ok to come this way.» So when this «vacuum» arrives at the valve, it actually helps emptying the cylinder. If they arrive and the valve is close (whether weak or strong pressure waves), they just bounce back into the pipe with the same strength as they came, kind of saying: «There is no way thru here, don't come in here.» These pressure waves bounce back like that until the pressure equalizes across the pipe. Of course, because in an engine the exhaust valve keeps opening with fresh new pressurized gases, it never stabilizes. So the pressure waves are everywhere, crossing path with each other. Once in the collector, they go back into the other primary pipes (they look for an exit everywhere) and once they go back into the actual outlet pipe, once they reach another area change (cat, muffler, atmosphere. etc), they send back another weak pressure wave that will go back into every primary pipe (even weaker as it is divided by number of primary pipes). You can see the whole mathematical puzzle of keeping track of all of those pressure waves. And I haven't even got into the math, as when they cross, their pressure combines, which affects their local temperature, which affect their local speed of sound, which means that their actual speed slows down or speeds up while they cross. A big mess. Simplification primary pipe area What we know is that the smaller the pipe area, the stronger will be the pressure wave. But if the area is too small, the actual flow will choke. So it has been found that the port area that creates average port velocities of about 75 m/s performs best. So the volumetric flow rate ˙ V V˙ is equal to the port velocity v p vp times the port area A p Ap: ˙ V = v p A p V˙=vpAp But the volumetric flow rate is also dependent on what the engine produces which is one cylinder volume V c y l Vcyl per exhaust event duration θ θ at the given rpm ω ω: ˙ V = V c y l ω θ V˙=Vcylωθ So: ω = A p θ v p V c y l ω=ApθvpVcyl This is in SI units, so to convert angular velocity from rpm to rad/s, area from in² to m², angle duration from deg to rad and volume from in³ to m³, we have to include the following conversion factor: ω = 6.5617 A p θ v p V c y l ω=6.5617ApθvpVcyl If we assume that the exhaust duration is one stroke, then θ θ = 180° and that v p vp = 75 m/s then: 6.5617 ( 180 ) ( 75 ) = 88 583 6.5617(180)(75)=88583 or: ω = 88 583 A p V c y l ω=88583ApVcyl Which looks a lot like your equation. You referred to A. Graham Bell. His equation for the diameter of the exhaust pipe D p Dp, based on exhaust length L L, is: D p = 2.1 √ V c y l 25 ( L + 3 ) Dp=2.1Vcyl25(L+3) Putting it for the area A p Ap instead: A p = π 4 D 2 p = π 4 2.1 2 V c y l 25 ( L + 3 ) Ap=π4Dp2=π42.12Vcyl25(L+3) Replacing L L by his other equation you mentioned, we get: A p = 3.46 V c y l 25 ( 850 θ ω ) = V c y l ω 6141.6 θ Ap=3.46Vcyl25(850θω)=Vcylω6141.6θ But in his equation, V c y l Vcyl is in cm³ so it has to be converted to in³ to be consistent with our other equation, so it becomes: A p = V c y l ω 374.8 θ Ap=Vcylω374.8θ or: ω = 374.8 A p θ V c y l ω=374.8ApθVcyl Comparing to our theoretical equation (in red) 374.8 = 6.5617 v p 374.8=6.5617vp or v p vp = 57.11 m/s. This might seem different than the 75 m/s of the previous equation, but Bell doesn't assume the exhaust duration is 180°, he takes the 180° + the exhaust valve opening duration. If we assume an exhaust valve opening of 55°, then θ θ = 180 + 55: 374.8 ∗ ( 180 + 55 ) = 88 078 374.8∗(180+55)=88078 Which is close to the 88 200 that you find in the equation you gave in the OP. Conclusion: v p vp is found empirically. Bell took care of including a better approximation of exhaust duration in his equation to find it, so it might be a better assumption. But it is far from an exact science. Collector area You mentioned that the «multiplication factor for collector size is anywhere from 1.6 - 1.8». Where does it come from? I told you that it is the area change that counts when comes the time to reflect the strong pressure wave as a weak pressure wave. The largest area change is when the pipe exit directly to the atmosphere (straight pipe), in which case the area ratio is infinite. But it has been found that when the area ratio is larger than about 6, the effect increases very little, meaning that it is almost like being straight pipe. Here is how we calculate the collector's area ( A c Ac) vs the primary pipe's area( A p Ap). The primary pipe exits into the collector pipe, but also in the other primary pipes (say there are N N primary pipes in total), then if we want the area ratio to be greater than 6: 6 < A c + ( N − 1 ) A p A p 6 7 − N AcAp>7−N So, if N N = 4, then A c A p > 3 AcAp>3 and since the area is proportional to the square of the diameter, D c d p > √ 3 Dcdp>3 or 1.73. That is where 1.6-1.8 comes from. You have a V6 - so your collector have 3 primary pipes, not 4 - thus a diameter ratio of 2 would be best suited. Primary pipe length To define the length of the pipe, we know that a pressure wave has to travel the entire length of the primary pipe L L, then comes back to the valve and all of that at the speed of sound v s vs, so the time t t taken to do that is: t = 2 L v s t=2Lvs We usually want it to return when the valve closes to help empty the cylinder. The time between the opening of the exhaust valve (when the pressure begins its journey down the pipe) and the end of the exhaust stroke can be found with the crank angle duration θ θ and the rpm ω ω: t = θ ω t=θω Such that: 2 L v s = θ ω 2Lvs=θω Or: L = v s θ 2 ω L=vsθ2ω Those are in SI units, so converting rad/s to rpm, rad to deg and m to in you have to include the following conversion factor: L = 3.28 v s θ ω L=3.28vsθω you should already see a resemblance with Bell's equation: L = 850 θ ω − 3 L=850θω−3 Where θ θ is 180° + the exhaust valve opening duration. We already mentioned that the speed of sound varies a lot from one end of the pipe to the other, so we can't use the true speed of sound. But we can used some sort of «average» speed of sound for the pipe. This «average» speed is found empirically by testing different pipe length on the same engine. Bell's equation has 850 = 3.28 v s 850=3.28vs or v s vs = 259 m/s. Bell just subtracted 3 in from this calculated length to take into account the exhaust port length. Of course, you can adjust this number for your particular engine, but this equation is such a crude approximation that it probably wouldn't matter much anyway. Collector length For the collector length, if you have an area ratio greater than 6, it doesn't matter much: For the exhaust gases, it will feel like they reached the atmosphere. The greater the area ratio, the less significant is the length. Still, in one of my books, there is a «large ballpark» figure for a high-speed drag-racing engine which says that the collector's length should be half the primary pipe length. If the primary pipe length is a crude approximation, imagine how worst is this one. With a stock vehicle - with cat, muffler and exit pipe - these would also have an effect, but very little if you've managed to respect the area ratio greater than 6.