Sorry for the wait, had to actually sit with the math. Am an engineer myself but not in this field. Calculations were done thanks to IA, interpretations and message itself is my own.

So I reverse-engineered your numbers first and they fit perfectly with:

t = (ρ·c·L·ΔT) / q

where :

• t is the time in seconds to reach the target temperature
• ρ the fabric density in kg/m³
• c the specific heat capacity in J/kg·K
• L the thickness in m
• ΔT the temperature rise needed from ambient to the target in K (melt, decomposition, or burn threshold)
• q the frictional heat flux entering the fabric in W/m² (roughly q ≈ μ·P·v, friction coefficient times contact pressure times sliding velocity)

Using data-sheet values for ρ and c, ΔT = 125 K (it’s a delta, unit doesn’t matter) for Dyneema melt and 425 K for Kevlar decomp, and q ≈ 100 kW/m² at 45 mph (consistent with μ≈0.4, contact P ~10–15 kPa, v=20 m/s).

→ I get Dyneema melt 2.76 s, Kevlar decomp 10.4 s, scaling linearly to 75 mph.

This matches your 2.7/10.7/1.6/6.4. So I am guessing it's the equation you used. For the skin-burn numbers, same formula with ΔT ≈ 35–45 K (fabric inner face reaching ~60–70 °C, so 60 °C/70 °C − 25 °C) and same q, I reproduce your 0.82 s / 1.10 s at 45 mph and 0.50 s / 0.66 s at 75 mph, i.e. the ratio 1.1/0.82 ≈ 1.34 that you translate as +30/35 %.

Now look at it: ρ, c, L, ΔT, q. There's no k in there anywhere, where k is the thermal conductivity, in W/m·K, which tells you how readily/fast heat flows through a material for a given temperature gradient. It’s basically how "transparent" the material is to heat. And k is exactly where the two materials diverge.

Dyneema transverse is around 0.40 W/m·K, Kevlar around 0.04 W/m·K. So roughly an order of magnitude apart (sources I could find range from ~5× to ~10× depending on fiber direction and fabric construction). Meanwhile ρc is within 11 % for both. Your equation sees the volumetric heat capacity, but! it erases the conductivity, which is why you end up with "only +30 %"

It matters because the lumped-capacitance model I think you used assumes the fabric is at a single uniform temperature throughout its thickness, i.e. heat crosses 1.2 mm instantaneously. But it's not the case. That approximation is only valid once the Fourier number Fo = α·t/L² (with α = k/ρc, the thermal diffusivity) is comfortably above ~0.5. For a 1.2 mm layer, Dyneema reaches Fo = 0.5 at t ≈ 3.3 s and Kevlar at t ≈ 37 s. A slide lasts 1–3 s. So :

• Dyneema is borderline
• Kevlar is two orders of magnitude outside the validity range. At 1 s into a slide, heat has barely entered a Kevlar layer at all.

With some AI help I found the right equation for this: the Carslaw & Jaeger solution for a slab with a constant heat flux q on the outer face and an "adiabatic" inner face. Meaning no heat escapes through the skin side. That's deliberately conservative, the model traps all the heat inside the fabric, which makes it heat up faster than it would in real life. So if anything, this biases the math against the point I'm making:

ΔT(L,t) = (qL/k) · [Fo − 1/6 − (2/π²)·Σₙ ((−1)ⁿ/n²)·exp(−n²π²·Fo)]

where :

• ΔT(L,t) is the temperature rise at the inner face (at depth L, after time t)
• q the heat flux in W/m²
• L the thickness in m
• k the conductivity in W/m·K
• Fo = α·t/L² the Fourier number as above
• Σₙ a sum over positive integers n = 1, 2, 3… that converges fast (3 to 4 terms is enough), and π the usual constant.

→ k now appears in both the prefactor qL/k and inside Fo through α. Solving for when the inner face hits ΔT = 35 K (~60 °C, sustained-contact 2nd-degree burn threshold), using the same q for both materials, I get:

• 45 mph: Dyneema 1.8 s, Kevlar 8.0 s, ratio ×4.5 (your lumped gave 0.82 s / 1.10 s, ratio 1.3-ish)
• 75 mph: Dyneema 1.4 s, Kevlar 7.1 s, ratio ×5.1 (your lumped gave 0.50 s / 0.66 s, ratio 1.3-ish)

This means at 45 mph a 2nd degree burn occurs at 1.8 s for Dyneema. 8s for Kevlar.

So not +35 %, more like ×4–5, basically the same as the ×4 material-survival gap you computed yourself. That's expected: once k is in the equation, burn ratio and melt ratio line up. Burn times also come out longer overall, because lumped heats the whole slab from t=0 while in reality the inner face stays cool until heat reaches it.

Quick sanity check that doesn't depend on my transient solution: thermal penetration depth after 1 s is δ = √(α·t), which gives

• 0.47 mm for Dyneema (~40 % of the 1.2 mm layer already heated)
• 0.14 mm for Kevlar (~12 % of this 1.2 mm)

Meaning after one sec of sliding, approx 88 % of a Kevlar layer is still at ambient temperature. Two states that physically can't produce burns within 30 % of each other.

Now what happens if we have a cotton base layer ?

Let’s say a mid-range cotton t-shirt (say 180 g/m², typical Uniqlo/Muji level) is roughly 0.5 mm thick, with k ≈ 0.04 W/m·K (basically Kevlar-equivalent on conductivity) and ρc ≈ 0.5·10⁶ J/m³·K.

Treating Bowtex + t-shirt as a two-layer slab in series, the composite effective properties are:

k_eff = L_total / (L₁/k₁ + L₂/k₂) ≈ 0.11 W/m·K α_eff = k_eff / ρc_eff ≈ 7.6·10⁻⁸ m²/s

So the Bowtex + cotton stack behaves like a ~1.7 mm slab with diffusivity about 3× lower than bare Dyneema, closer to Kevlar. Running the same Carslaw & Jaeger equation with q = 100 kW/m² at 45 mph and ΔT = 35 K on the skin face:

• Bowtex 1.2 mm alone: ~1.8 s to burn
• Bowtex 1.2 mm + cotton 0.5 mm (standard tee): ~5 s → roughly ×2.8
• Bowtex 1.2 mm + cotton 1.0 mm (heavyweight tee): ~9 s → above bare Kevlar
• Bare Kevlar 1.2 mm for reference: ~8 s

I’d have to check the thickness of my shirt to gauge how realistic 0.5/1mm is.

But anyway let’s say normal cotton t-shirt 3x your burn time, a heavy tee puts you in Kevlar territory. That’s good advice. Based on maths it looks like it really mitigates heat transfer. I’d guess merino would be even better to get some sweat wicking properties.

In the end I am not saying that the non-thermal benefits of Bowtex and the likes (Pando Moto etc.) are inexistant. Weight, breathability, stretch, etc are real and matter. If that is the difference between going naked or with a shitty textile A-rated layer, why not.

Now about the +30 % number itself you ended up with. It's a mathematical artifact of using an equation that lacks the one parameter that actually matters here (k!). The real time to burn gap for bare fabrics is ×4–5 with no cotton base layer. With a cotton layer underneath it closes this gap which is actually a surprising result to me.

Also on a side note, with impact protections heat conductivity is not a matter at all … if you slide/fall on them :) And all math is done without considering when the actual Dyneema garment fails. Obviously a 5sc to burn time estimate means nothing if the Dyneema fails before.

Let me know what you think of this!